First post

Proof. The proof is a little technical. We need the following lemma:

Now apply the lemma for the function $$\phi(g, t) = (D\rho(\exp(tX))f - \rho(\exp(tX))Df)(g)$$ and we get the result. Note that $G$ is generated by $\exp(\mathfrak{g})$. ◻

Proof. This follows from direct computations and relations among $\widehat{R}, \widehat{L}, \widehat{H}$. ◻

Proof. For 1, we can check that $\pi(X)\pi(\phi)f = \pi(\phi_{X})f$ where $$\phi_{X}(g) = \frac{d}{dt}\Big|_{t=0} \phi(\exp(-tX)g).$$ Hence $\pi(\phi)f$ is differentiable and we can repeat this to get $\pi(\phi)f\in \mathfrak{H}^{\infty}$.

For 2, we use 1 with appropriate function $\phi$. For given $\epsilon>0$, continuity of $(g, f)\to \pi(g)f$ implies that there exists an open neighborhood of the identity of $G$ such that $|\pi(g)f - f|<\epsilon$ for all $g\in U$. Now take $\phi\in C_{c}^{\infty}(G)$ to be a nonnegative function with $\mathrm{supp}\,(\phi)\subset U$ and $\int_{G}\phi(g) dg = 1$, so that $$|\pi(\phi)f - f| \leq \int_{G} \phi(g)|\pi(g)f - f|dg \leq \epsilon$$ which proves that $\mathfrak{H}^{\infty}$ is dense in $\mathfrak{H}$. ◻

Proof. We define such inner product on $\mathfrak{H}$ as $$\langle v, w\rangle = \int_{K} \langle \pi(\kappa)v, \pi(\kappa)w\rangle_{1}d\kappa.$$ It is easy to check that this defines a new inner product which is $K$-invariant. By Banach-Steinhaus theorem, we can found a constant $C>0$ such that $C^{-1}|v|_{1} \leq |\pi(\kappa)v|_{1} \leq C|v|_{1}$ for all $v\in \mathfrak{H}$ and $\kappa \in K$, and this proves $C^{-1}|v|_{1} \leq |v| \leq C|v|_{1}$ for all $v$. Hence topologies are same. ◻

Proof. Assume that $f_{i} = \langle \pi_{i}(g)x_{i}, y_{i}\rangle$ such that $$\overline{\langle\langle f_{1}, f_{2}\rangle\rangle} := \int_{G} \overline{f_{1}(g)} f_{2}(g) dg = \int_{G} \overline{\langle \pi_{1}(g)x_{1}, y_{1}\rangle}\langle \pi_{2}(g)x_{2}, y_{2}\rangle dg \neq 0.$$ Then the bounded linear map $L:H_{1}\to H_{2}$ defined as $$L(v) = \int_{G} \langle \pi_{1}(g)v, y_{1}\rangle \pi_{2}(g^{-1})y_{2} dg$$ gives a nonzero intertwining operator, since $\langle x_{2}, L(x_{1})\rangle = \overline{\langle\langle f_{1}, f_{2}\rangle\rangle}$. ◻

Proof. By embedding $\mathrm{GL}(n, \mathbb{C})\hookrightarrow \mathrm{GL}(2n, \mathbb{R})$, we can assume that $K$ is a subgroup of $\mathrm{GL}(n, \mathbb{R})$ for some $n$. We call a function on $K$ a polynomial function if it sis a polynomial with complex coefficients in terms of $n^{2}$ entries of matrices in $K\subset \mathrm{Mat}(n, \mathbb{R})$. We first show that any polynomial function on $K$ is a matrix coefficient of a finite dimensional representation. Indeed, let $r\in \mathbb{Z}_{>0}$ and $(\rho, R)$ be the representation of $K$ where $R$ is a space of polynomial functions of degree $\leq r$ on $\mathrm{Mat}(n, \mathbb{R})$, where $K$ acts by right translation. We can find a Hiermitian inner product on $R$ which is $K$-invariant, and by Riesz representation theorem there exists $f_{0}\in R$ such that $f(1) = \langle f, f_{0}\rangle$ for all $f\in R$, since $f\mapsto f(1)$ is a bounded linear functional on $R$. Then $$f(g) = (\rho(g)f)(1) = \langle \rho(g)f, f_{0}\rangle$$ so the function $f$ is a matrix coefficient of $R$.

Now we prove 1. It is known that $C(K)$ is dense in $L^{p}(K)$ for any $1\leq p <\infty$, and Stone-Weierstrass theorem implies that any continuous function on $K$ can be uniformly approximated by polynomial functions, which are matrix coefficients.

To show 2 and 3, it is enough to show that any nonzero unitary representation $(\pi, \mathfrak{H})$ of $K$ admits a nonzero finite dimensional invariant subspace. Choose any nonzero matrix coefficient $\phi$ of $\mathfrak{H}$ and approximate it by a polynomial function $\phi_{0}$, so that $\phi$ and $\phi_{0}$ are not orthogonal. Then the proposition \ref{nonzeroint} shows that there is a nonzero intertwining map $L:R\to \mathfrak{H}$ for a finite dimensional representation $R$ of polynomial functions, and the image of $L$ is a finite dimensional invariant subspace of $\mathfrak{H}$. This proves 2, and 3 also follows from this with applying Zorn’s lemma. ◻

Proof. We need the following decomposition theorem of Cartan, which we will not going to prove. Basically, this follows from the induction on $n$.

Now let $\iota:C^{\infty}_{c}(K\backslash G / K)\to C^{\infty}_{c}(K\backslash G /K)$ be a map defined as $\iota(\phi(g))= \widehat{\phi}(g) := \phi(\prescript{T}{}{g})$. Then this map ins an anti-involution of $C^{\infty}_{c}(K\backslash G / K)$: \begin{align*}\widehat{(\phi_{1}*\phi_{2})}(g) &= \int_{G} \phi_{1}(\prescript{T}{}{}gh)\phi_{2}(h^{-1})dh \\ &= \int_{G} \widehat{\phi_{2}}({}^{T}h^{-1}) \widehat{\phi_{1}}({}^{T}hg) dh \\ &= \int_{G} \widehat{\phi_{2}}(h) \widehat{\phi_{1}}(h^{-1}g)dh = (\widehat{\phi_{2}} * \widehat{\phi_{1}})(g). \end{align*} By the way, Cartan’s decomposition theorem allow us to decompose $g$ as $g = \kappa_{1} d\kappa_{2}$ where $\kappa_{1}, \kappa_{2}\in K$ and $d$ is a diagonal matrix. Then $\phi(g) = \phi(d) = \widehat{\phi}(d) = \widehat{\phi}(g)$, so that $\iota = \mathrm{id}$ and $\phi_{1} * \phi_{2} = \phi_{2} * \phi_{1}$, i.e. $C_{c}^{\infty}(K\backslash G /K)$ is commutative. ◻

Proof. The proof is almost same, but we use the following involution $$\widehat{\phi}(g) = \phi\left(\begin{pmatrix} -1 & \\ & 1 \end{pmatrix} \prescript{T}{}{}g \begin{pmatrix} -1 & \\ & 1 \end{pmatrix}\right).$$ ◻

Proof. By admissibility, we know that $\mathfrak{H}^{K}$ is finite dimensional. $C_{c}^{\infty}(K\backslash G/K)$ can be realized as a commutative family of normal operators on a finite dimensional space, which are simultaneously diagonalizable. Therefore there is a one dimensional invariant subspace $V_{0}$ of $\mathfrak{H}^{K}$, which should be whole $\mathfrak{H}^{K}$ by irreducibility. The proof is almost same for $\mathfrak{H}_{k}$ except that we use commutativity of $C_{c}^{\infty}(K\backslash G/K, \sigma_{k})$ instead of $C_{c}^{\infty}(K\backslash G/K)$. ◻

Proof. Let $\mathfrak{H}_{0} = \mathfrak{H}^{\infty} \cap \mathfrak{H}_{\mathrm{fin}}$. We will first show that $\mathfrak{H}_{0}$ is dense in $\mathfrak{H}^{\infty}$. For given $f\in \mathfrak{H}$, we will find suitable $\phi \in C^{\infty}(G)$ such that $\pi(\phi)f$ is sufficiently close to $f$ and $\pi(\phi)f \in \mathfrak{H}_{0}$. To do this, let $U$ be a small open neighborhood of the identity in $G$ and let $\epsilon >0$ be a given constant. Choose $U_{1} \subset U$ and $V\subset K$ such that $VU_{1} \subset U$. Let $\phi_1$ be a smooth positive-valued function with $\mathrm{supp}\,(\phi_1)\subset U_1$ and $\int_{F} \phi_{1}(g) dg = 1$. Also, by Peter-Weyl theorem, we can find a matrix coefficient $\phi_0$ of a finite dimensionalunitary representation $(\rho, R)$ of $K$ such that $\int_{K} \phi_0(\kappa)d\kappa = 1$ and $\int_{K\backslash V} |\phi_{0}(\kappa)| d\kappa < \epsilon$. Now let $$\phi(g) := \int\displaylimits_{K} \phi_{0}(\kappa) \phi_{1}(\kappa^{-1}g)d\kappa.$$ Then one can check that $\int_{G\backslash U} |\phi(g)|dg < \epsilon$, so that $\pi(\phi)f$ is sufficiently close to $f$. To show that $\pi(\phi)f$ is $K$-finite, let $\phi_{0}(\kappa) = \langle \rho(\kappa)\xi, \eta\rangle$ where $\xi, \eta$ are vectors in $R$. Then for $\kappa_1\in K$, we have \begin{align*}\phi_{1}(\kappa^{-1}g) = \int\displaylimits_{K} \langle\langle \rho(\kappa)\xi, \rho(\kappa_1)\eta\rangle\rangle \phi_1(\kappa^{-1}g)d\kappa\end{align*} so the space of functions $\phi(\kappa_{1}^{-1}g)$ lies in the finite dimensionalspace spanned by functions of the form $$g\mapsto \int\displaylimits_{K} \langle\langle \rho(\kappa)\xi, \zeta\rangle\rangle \phi_{1}(\kappa^{-1}g)d\kappa, \quad \zeta\in R.$$ This is a finite dimensionalspace of functions, so the space spanned by the vectors $$\pi(\kappa_1) \pi(\phi)f = \int\displaylimits_{G}\phi(g)\pi(\kappa_{1}g)fdg = \int\displaylimits_{G} \phi(\kappa_{1}^{-1}g)\pi(g)fdg$$ is finite dimensional. Hence $\pi(\phi)f\in \mathfrak{H}_{\mathrm{fin}}$ by the previous proposition. This shows $\mathfrak{H}_0$ is dense in $\mathfrak{H}$.

To show $\mathfrak{H}_\mathrm{fin}\subseteq \mathfrak{H}^{\infty}$, it is enough to show that $\mathfrak{H}_{0}(\sigma) = \mathfrak{H}(\sigma)$ for all irreducible representation $\sigma$ of $K$. Clearly, $\mathfrak{H}_{0}(\sigma) \subseteq \mathfrak{H}(\sigma)$, and if they are not same for some $\sigma$, then we can find $0\neq f\in \mathfrak{H}(\sigma)$ orthogonal to $\mathfrak{H}_{0}(\sigma)$, Then this $f$ is orthogonal to $\mathfrak{H}_{0}(\tau)$ for all $\tau \neq \sigma$, which contradicts to the denseness of $\mathfrak{H}_{0}$ in $\mathfrak{H}^{\infty}$.

For $\mathfrak{g}$-invariance, let $f\in R\subset \mathfrak{H}$ be a $K$-finite vector where $R$ is a finite dimensional $\mathfrak{k}$-invariant subspace. Let $R_{1}$ be a space generated by $Yf$ for $Y\in \mathfrak{g}$ and $f\in R$, which is also a finite dimensional space. For $X\in \mathfrak{k}$ and $Y\in \mathfrak{g}$, $X(Y\phi) = [X, Y]\phi + Y(X\phi)$ shows that $R_{1}$ is $\mathfrak{k}$-invariant so $Yf$ is a $K$-finite vector. ◻

Proof. We can naturally extend the adjoint action $\mathrm{Ad}$ of $G$ on $\mathfrak{g}$ to $U\mathfrak{g}_{\mathbb{C}}$ by $$\mathrm{Ad}(g) (x_{1}\otimes \cdots \otimes x_{r}) = \mathrm{Ad}(g)x_{1} \otimes \cdots \otimes \mathrm{Ad}(g)x_{r}.$$ One can check that $D$ is fixed by this action by the third condition of $(\mathfrak{g}, K)$-module, so that $\pi(\kappa)\circ D = D\circ \pi(\kappa)$ for $\kappa\in K$. Consequently, the isotypic subspaces $V(\sigma)$ are stable under $D$. Choose any nonzero $V(\sigma)$. Since it has finite dimension, there exists a nonzero eigenvector $x_{0}\in V(\sigma)$ with an eigenvalue $\lambda$. Let $V_{0}\subseteq V(\sigma)$ be an eigenspace of $\lambda$. Since $D$ is in the center, it commutes with the action of $\mathfrak{g}$ and $K$, so that $V_{0}$ is a nonzero invariant subspace. Thus we have $V = V(\sigma) = V_{0}$. ◻

Proof. Every statement follows form the relations among $R, L, H$ and $\Delta$. ◻

Proof. Basically, all of these follows from the previous proposition, 6 and 7. For the uniqueness, we will only show the first case. Let $V, V'$ be two irreducible admissible $(\mathfrak{g}, K)$-module with the same set of $K$-types. Choose $0\neq x\in V(k)$ and $0\neq x'\in V'(k)$, then $x, L^{n}x, R^{n}x$ (for $n>0$) form a basis of $V$, and similarly $x', L^{n}x', R^{n}x'$ ($n>0$) form a basis of $V'$. Now if we define $\phi:V\to V'$ by $\phi(x) = x', \phi(L^{n}x) = L^{n}x'$ and $\phi(R^{n}x) = R^{n}x'$, then we can easily check that this is a nonzero $(\mathfrak{g}, K)$-module homomorphism from $V$ to $V'$. ◻

Proof. It is easy to check that the function $g\mapsto \langle f_{1}(g), f_{2}(g)\rangle$ is in $C(P\backslash G, \delta)$, i.e. satisfies $f(pg) =\delta(p)f(g)$ for all $p\in P$ and $g\in G$. One can prove that the linear functional $I:C(P\backslash G, \delta)\to \mathbb{C}$ defined as $I(f) = \int_{K} f(\kappa)d\kappa$ is $G$-invariant under the right regular representation, by showing that the map $\Lambda:C_{c}(G)\to C(P\backslash G, \delta), \phi\mapsto (g\mapsto \int_{P} \phi(pg)dp)$, is surjective and $I(\Lambda f) = \int_{G}f(g)dg$. For details, see Lemma 2.6.1 in (bu?). ◻

Proof. With the assumption, we can easily check that $s_{1}, s_{2}$ satisfying $\mu = s_{1} + s_{2}, s = \frac{1}{2}(s_{1}- s_{2} + 1), \lambda = s(1-s)$ are all pure imaginary. Then the character $\chi:B(\mathbb{R})^{+}\to \mathbb{C}^{\times}$ defined as $$\chi\begin{pmatrix} y_{1} & x \\ & y_{2} \end{pmatrix} = \mathrm{sgn}(y_{1})^{\epsilon}|y_{1}|^{s_{1}}|y_{2}|^{s_{2}}$$ is unitary and the induced representation that is contained in the class $\mathcal{P}_{\mu}(\lambda, \epsilon)$ is also unitary by the previous theorem. ◻

Proof. 1 follows from the fact that $Z\in \mathfrak{g}$, so action is skew-symmetric, and the action of $\Delta$ is symmetric. For 2, we know that $\mathfrak{H}(k)\neq 0$ for all $k\equiv\epsilon\,(\mathrm{mod}\,2)$. From $-4\Delta - H^{2} + 2H = 4RL$, $Hf_{k} = kf_{k}$ (where $0\neq f_{k}\in \mathfrak{H}(k)$), and $\langle RL f_{\epsilon}, f_{\epsilon}\rangle = \langle Lf_{\epsilon}, Lf_{\epsilon}\rangle >0$, we get $-4\lambda - \epsilon^{2} + 2\epsilon <0$ which gives the results. ◻

Proof. It is almost direct to check that the map is indeed an intertwining map, if we know that the intertwining map is convergent. For the convergence, we only need to check convergence for $g = 1$ (since $M(s)$ is an intertwining map). The identity \begin{align*}\begin{pmatrix} & -1 \\ 1 & \end{pmatrix}\begin{pmatrix} 1 & x \\ & 1 \end{pmatrix} &= \begin{pmatrix} \Delta_{x}^{-1} & -x\Delta_{x}^{-1} \\ & \Delta_{x} \end{pmatrix} \kappa_{\theta(x)} \\ \Delta_{x} &= \sqrt{1+x^{2}}, \quad \theta(x) = \arctan\left(-\frac{1}{x}\right). \end{align*} gives $$(M(s)f)(1) = \int_{-\infty}^{\infty} (1+x^{2})^{-s}f(\kappa_{\theta(x)})dx,$$ and by the boundedness of $f$ on $K$, the integral converges if $$\int_{-\infty}^{\infty} \frac{1}{|(1+x^{2})^{s}|} dx$$ converges, which is true for $\Re s>\frac{1}{2}$. To check that $M(s)f\in H^{\infty}(s_{2}, s_{1}, \epsilon)$, it is enough to check the following equations \begin{align*}(M(s)f)\left(\begin{pmatrix} 1 & \xi \\ & 1 \end{pmatrix}g\right) &= (M(s)f)(g) \\ (M(s)f)\left(\begin{pmatrix} y_{1} & \\ & y_{2} \end{pmatrix}g\right) &= |y_{1}|^{s_{2} + \frac{1}{2}} |y_{2}|^{s_{1} - \frac{1}{2}} (M(s)f)(g)\end{align*} for $\xi\in \mathbb{R}$ and $y_{1}, y_{2}>0$, which can be checked by direct computation (with some substitutions). Smoothness and $K$-finiteness are also can be easily checked from $$(M(s)f)(\kappa_{t}) = \int_{-\infty}^{\infty} (1+x^{2})^{-s}f(\kappa_{\theta(x)+t})dx.$$ ◻

Proof. It is enough to show for $g = 1$, which is equivalent to $$\int_{-\infty}^{\infty} (1+x^{2})^{-s}\exp(ik\theta(x)) dx = (-1)^{k} \sqrt{\pi} \frac{\Gamma(s)\Gamma\left( s-\frac{1}{2}\right)}{\Gamma\left(s+\frac{k}{2}\right)\Gamma\left(s-\frac{k}{2}\right)}.$$ Under the substitution $y = \frac{x-i}{x+i}$, the integral equals $$2i(-i)^{k} 4^{-s} \int_{C} (1-y)^{2s-2}(-y)^{\frac{k}{2}-s}dy$$ where $C$ is a contour consisting of unit circle centered at the origin and moves counterclockwise. For the convergence, we may assume $\Re(2s - 1), \Re(\frac{k}{2}-s) >0$, and use analytic continuation on $k$. If we deform the contour $C$ so that it proceeds directly from 1 to 0 along real axis, circles the origin in the counterclockwise direction, then returns to 1 along the real line, then the integral became \begin{align*}2i&(-i)^{k}4^{-s}[e^{-i\pi(s-k/2)} - e^{i\pi (s-k/2)}] \int_{0}^{1}(1-y)^{2s-2}y^{k/2-s}dy \\ &=(-1)^{k}\sqrt{\pi} \frac{\Gamma(s)\Gamma\left( s- \frac{1}{2}\right)}{\Gamma\left(s+\frac{k}{2}\right)\Gamma\left(s-\frac{k}{2}\right)},\end{align*} which follows from the Beta function identity and some other formulas of Gamma function. ◻

Proof. For $s_{1}, s_{2}\in \mathbb{C}$, we have a Hermitian pairing \begin{align*}H^{\infty}(s_{1}, s_{2}, \epsilon) &\times H^{\infty}(-\overline{s_{1}}, -\overline{s_{2}}, \epsilon) \to \mathbb{C}\\ (f, f')&\mapsto \int_{K} f(\kappa) \overline{f'(\kappa)} d\kappa\end{align*} which is $G$-invariant.

Now assume that $s_{1} = -\overline{s_{2}}$ and $s_{2} = -\overline{s_{1}}$, so that $\mu = s_{1} + s_{2} \in i\mathbb{R}$ and $s = \frac{1}{2}(s_{1} - s_{2} + 1)\in \mathbb{R}$. Since $H^{\infty}(s_{2}, s_{1}, \epsilon) = H^{\infty} (-\overline{s_{1}}, -\overline{s_{2}}, \epsilon)$, we can define Hermitian pairing on $H^{\infty}(s_{1}, s_{2}, \epsilon)$ by $$\langle f, f'\rangle = \int_{K} f(\kappa)\overline{i^{\epsilon}(M(s)f')(\kappa)}d\kappa,$$ which is $G$-invariant. We only need to show that this pairing is positive definite, and it follows from the following computation $$\langle f_{k, s}, f_{k, s}\rangle = (-1)^{\frac{k}{2}} \sqrt{\pi} \frac{\Gamma(s)\Gamma\left(s-\frac{1}{2}\right)}{\Gamma\left(s+\frac{k}{2}\right)\Gamma\left(s-\frac{k}{2}\right)}$$ which is positive for $\frac{1}{2} <s < 1$ and even $k$. ◻

Proof. Finite dimensional unitary representation of $\mathrm{GL}(n, \mathbb{R})^{+}$ can be regarded as a homomorphism $\pi:\mathrm{GL}(n, \mathbb{R})^{+}\to U(m)$ where $m$ is the dimension of the representation. Since $U(m)$ is compact, image of $\pi$ is also compact. It is known that $\mathrm{SL}(n, \mathbb{R})$ is simple for odd $n$ and $\mathrm{PSL}(n, \mathbb{R}) = \mathrm{SL}(n, \mathbb{R})/\{\pm I\}$ is simple for even $n$, so the only compact homomorphic image of $\mathrm{SL}(n, \mathbb{R})$ is the trivial group. Hence $\mathrm{SL}(n, \mathbb{R})\subset \ker\pi$ and the representation factors through the determinant map. Now we know that the only unitary representation of $\mathbb{R}_{+}^{\times}$ are of the form $t\mapsto t^{r}$ for $r\in i\mathbb{R}$. ◻

Proof. The automorphism $\iota:G\to G$ defined as $$\iota\left(\begin{pmatrix} a & b \\ c & d \end{pmatrix}\right)= \begin{pmatrix} a & -b \\ -c & d \end{pmatrix}$$ relates $\pi^{+}$ and $\pi^{-}$ by $\pi^{+}(g)= \pi^{-}(\iota(g))$. Thus it is sufficient to show that $(\pi^{-}, \mathfrak{H})$ is an irreducible admissible representation in $\mathcal{D}^{-}(k)$.

Define the representation $(\pi, \mathfrak{H})$ by $$\pi(g) f= f|_{k}g^{-1}$$ where $|_{k}$ is the weight $k$ slash operator, i.e. $$(f|_{k}g)(z) = \det(g)^{k/2} (cz+d)^{-k} f\left( \frac{az+b}{cz+d}\right)$$ for $g\in \mathrm{GL}(2, \mathbb{C})$. Then $\pi\simeq \pi^{-}$ since $\pi^{-}(g)f = \pi(w_{0}gw_{0}^{-1})f$ for $w_{0} = \left(\begin{smallmatrix} 0 & -1 \\ 1 & 0 \end{smallmatrix}\right)$. So it is enough to show that $(\pi, \mathfrak{H})$ is an irreducible admissible representation in $\mathcal{D}^{-}(k)$.

Let $s_{1} = -s_{2} = (k-1)/2$, so that $s = k/2$ and $\mu = 0$, and let $\epsilon\in \{0, 1\}$ with $\epsilon\equiv k\,(\mathrm{mod}\,2)$. We can define a bilinear pairing $\langle\langle \,, \rangle\rangle: H^{\infty}(s_{1}, s_{2}, \epsilon) \times H^{\infty} (-s_{1}, -s_{2}, \epsilon) \to \mathbb{C}$ by $$\langle\langle f_{1}, f_{2}\rangle\rangle = \int_{K} f_{1}(\kappa)f_{2}(\kappa) d\kappa$$ which is $G$-equivariant. Now we define a map $\sigma:H^{\infty}(-s_{1}, -s_{2}, \epsilon)\to C^{\infty}(G)$ by $$(\sigma f)(g) = \langle\langle \rho(g)f_{k, s}, f\rangle\rangle,$$ then the function $$(\Sigma f)(z) = y^{-k/2}(\sigma f)\left(\begin{pmatrix} y^{1/2} & xy^{-1/2} \\ & y^{1/2} \end{pmatrix}\right)$$ is an holomorphic function on $\mathcal{H}$ that satisfies $(\Sigma f|_{k}g)(i) = (\sigma f)(g)$. (Holomorphicity follows from $L(\sigma f) =0$.)

We will now prove that $$\Sigma f_{l, 1-s} = c(l) (z-i)^{-(i+k)/2}(z+i)^{(l-k)/2}$$ for some constant $c(l)$ which is zero for $l>-k$. We have $$(\sigma f_{l, 1-s})(\kappa_{\theta}g) = \langle\langle \rho(g)f_{k, s}, \rho(\kappa_{\theta}^{-1})f_{l, 1-s}\rangle\rangle = e^{-il\theta}(\sigma f_{l, 1-s})(g),$$ and this implies that the function $\phi = \Sigma f_{l, 1-s}$ satisfies $$\phi|_{k}\kappa_{\theta} = e^{-il\theta}\phi.$$ If $C = -\frac{1+i}{2} \left(\begin{smallmatrix} i & 1 \\ i & -1 \end{smallmatrix}\right)$ (Cayley transform), $\psi:= \phi_{k}|C^{-1}$ is a function on the unit disk with $$\psi\Big|_{k} \begin{pmatrix} e^{i\theta} & \\ & e^{-i\theta} \end{pmatrix} = e^{-il\theta}\psi$$ and if we consider the Taylor expansion of $\psi$, we get $\psi(w) = cw^{(-l-k)/2}$ for some constant $c$, which implies $\phi= c(l) (z-i)^{-(i+k)/2}(z+i)^{(l-k)/2}$ where $c(l) =0$ for $l>-k$. From this, the kernel of the map $\Sigma:H(-s_{1}, -s_{2}, \epsilon) \to C^{\omega}(\mathcal{H})$ contains the (reducible) invariant subspace $\langle f_{l, 1-s}\,:\, l\geq 2-k\rangle$, and we can check that $\Sigma f_{l, 1-s}$ are all square-integrable for $l\leq -k$ by using the explicit description, so the image lies in $\mathfrak{H}$. Also, $\Sigma f_{l, 1-s}$ span $\mathfrak{H}$ for $l\leq -k$ because as a function on the unit disk (via Cayley transform), power series expansion of a holomorphic function on the unit disk can be regarded as a Fourier expansion in terms of $\Sigma f_{l, 1-s}$. This completes the proof of 1.

For 2, note that the correspondence between $\sigma f$ and $\Sigma f$ is an isometry, and this gives a realization of $\mathcal{D}^{-}(k)$ in the left regular representation of $G$ on $L^{2}(G/Z)$, and it can be transferred to the right regular representation by composing with $g\mapsto g^{-1}$. ◻

Proof. Let $(\pi, \mathfrak{H})$ and $(\pi', \mathfrak{H}')$ be irreducible admissible unitary representations of $G = \mathrm{GL}(2, \mathbb{R})^{+}$ such that the spaces $V = \mathfrak{H}_{\mathrm{fin}}$ and $V' = \mathfrak{H}_{\mathrm{fin}}'$ are isomorphic as $(\mathfrak{g}, K)$-modules, and let $\phi:V\to V'$ be an isomorphism. Decompose $V$ and $V'$ as $V = \oplus_{k} V(k), V' = \oplus_{k} V'(k)$ and choose $k$ so that $V(k)\neq 0$. Then we can find $0\neq x\in V(k)$ which satisfies $|x| = 1$, and by normalizing $\phi$ we can also assume that $|\phi(x)| = 1$. (Note that all the spaces $V(k)$ are at most 1-dimension.) Then $$|Rx|^{2} = \langle Rx, Rx\rangle = -\langle LRx, x\rangle = \left(\lambda + \frac{k}{2}\left(1+\frac{k}{2}\right)\right) \langle x, x\rangle = \lambda + \frac{k}{2}\left(1+\frac{k}{2}\right),$$ and we get the same result for $|R\phi(x)|$. By repeating this, we can prove that $|R^{n}x| = |R^{n}\phi(x)|$ and $|L^{n}x| = |L^{n}\phi(x)|$ for all $n\geq 1$, which proves that $\phi$ is an isometry. Since $\mathfrak{H}$ and $\mathfrak{H}'$ are Hilbert space completions of $V$ and $V'$, we can extend $\phi$ to an isometry $\phi:\mathfrak{H}\to \mathfrak{H}'$.

Now we have to show that $\phi$ is an intertwining operator. For $f\in V = \mathfrak{H}_{\mathrm{fin}}$ and $X\in \mathfrak{g}$, we have \begin{align*}\phi(\pi(e^{X})f) = \sum_{n\geq 0} \frac{1}{n!} \phi(X^{n}f) = \sum_{n\geq 0} \frac{1}{n!}X^{n}\phi(f) = \pi'(e^{X})\phi(f)\end{align*} and the result follows from the fact that $V\subset \mathfrak{H}$ is dense and $G$ is generated by elements of the form $e^{X}$. ◻

Proof. We will assume that $\psi(x) =\psi_{1/2}(x) = e^{ix/2}$. The condition $ZW = \mu W$ implies $$W\left(\begin{pmatrix} u & \\ & u \end{pmatrix}g\right) = u^{\mu} W(g)$$ and we get \begin{align*}W(g) = u^{\mu} e^{i(x/2+k\theta)}w(y), \quad w(y) = W\begin{pmatrix} y^{1/2} & \\ & y^{-1/2} \end{pmatrix}\end{align*} where $g = \left(\begin{smallmatrix} u & \\ & u \end{smallmatrix}\right) \left(\begin{smallmatrix} y^{1/2} & xy^{-1/2} \\ & y^{-1/2} \end{smallmatrix}\right) \kappa_{\theta}$. Then the condition $\Delta W = \lambda W$ is equivalent to the 2nd order differential equation $$w'' + \left(-\frac{1}{4} + \frac{k}{2y} + \frac{\lambda}{y^{2}}\right)w = 0.$$ It is known that there exists two linearly independent solutions of this equation, $W_{\frac{k}{2}, s-\frac{1}{2}}(y)$ and $W_{-\frac{k}{2}, s-\frac{1}{2}}(-y)$, which are asymptotically $e^{-y/2}y^{k/2}$ and $e^{y/2}(-y)^{-k/2}$. (Here $s = \frac{1}{2} + (-\lambda + \frac{1}{4})^{1/2}$, and such functions are classical Whittaker functions.) Thus the assumption of moderate growth excludes the second solution and $\mathcal{W}(\lambda, \mu, k)$ is 1-dimensional space spanned by the function $$W_{k, \lambda, \mu}(g) = u^{\mu}e^{i(x/2 + k\theta)} W_{\frac{k}{2}, s-\frac{1}{2}}(y),$$ which is known to be rapidly decreasing and analytic. The statement about $R$ and $L$ action also follows from analytic properties of classical Whittaker functions. ◻

Proof. Let $\mu, \lambda$ be scalars corresponds to action of $Z$ and $\Delta$. Decompose $V$ as $\oplus_{k} V(k)$. If $V(k)\neq 0$, then its image under the isomorphism with $\mathcal{W}(\pi, \psi)$ is $\mathcal{W}(\lambda, \mu, k)$, and the previous theorem implies uniqueness and analytic properties. ◻

Proof. Proof follows from direct computations. The inverse map is given by $$f(z) = F\left(\begin{pmatrix} y & x \\ & 1 \end{pmatrix}\right)$$ for given $F\in L^{2}(\Gamma \backslash G, \chi, k)$. ◻

Proof. First statement uses Zorn’s lemma. If we define $\Sigma$ to be the set of all sets $S$ of irreducible invariant subspaces of $L^{2}(\Gamma \backslash G, \chi)$ such that the elements of $S$ are mutually orthogonal, then there exists a maximal element $S$ in $\Sigma$. If we put $\mathfrak{H}$ as the orthogonal complement of the closure of the direct sum of the elements of $S$, then one can show that $\mathfrak{H}= 0$. (For details, see Theorem 2.3.3 in (bu?).)

For the second statement, it is equivalent to showing that $L^{2}(\Gamma \backslash G, \chi, k)$ decomposes into direct sum of eigenspaces of $\Delta$. In Proposition \ref{commchar}, we showed that $C_{c}^{\infty}(K\backslash G /K, \sigma)$ is a commutative ring, where $\sigma(\kappa_{\theta}) = e^{ik\theta}$. For each character $\xi$ of $C_{c}^{\infty}(K\backslash G/K, \sigma)$, let $H(\xi) := \{f\in L^{2}(\Gamma \backslash G, \chi, k)\,:\, \pi(\phi)f = \xi(\phi)f, \phi\in C_{c}^{\infty}(K\backslash G/K, \sigma)\}$. Here $C_{c}^{\infty}(K\backslash G/K, \sigma)\subset C_{c}^{\infty}(G)$ acts as $$\pi(\phi)f = \int\displaylimits_{G} \phi(g)\pi(g)f dg.$$ One can show that $$L^{2}(\Gamma \backslash G, \chi, k) = \bigoplus_{\xi} H(\xi),$$ where the direct sum is a Hilbert space direct sum and $\xi$ ranges through all distinct characters of $C_{c}^{\infty}(K\backslash G/K, \sigma)$ with $H(\xi)\neq 0$. Each of $H(\xi)$ is finite dimensional. (Most of the result follows from the spectral theorem for self-adjoint compact operators, applied to $\pi(\phi)$. See Theorem 2.3.4 of (bu?) for details.) Since $\Delta$ commutes with $\pi(\phi)$ (recall that $H$ lies in the center of $U\mathfrak{g}_{\mathbb{C}}$, and it is both invariant under left and right regular representations - see Theorem \ref{center}), the spaces $H(\xi)$ are $\Delta$-invariant, so these decomposes as a direct sum of $\Delta$-eigenspaces since $\Delta$ is self-adjoint. Hence $L^{2}(\Gamma \backslash G, \chi, k)$ also decomposes as $\Delta_k$-eigenspaces. ◻

Proof. The only point worth to mention is the connection between discrete series representations and holomorphic modular forms. The multiplicity of $\mathcal{D}^{+}(k)$ equals the dimension of the $\frac{k}{2}\left( 1- \frac{k}{2}\right)$-eigenspace in $L^{2}(\Gamma\backslash G, \chi, k)$, or in $L^{2}(\Gamma \backslash\mathcal{H}, \chi, k)$. This eigenspace is isomorphic to the space of modular forms $M_{k}(\Gamma, \chi)$: let $H$ be an irreducible subspace that is isomorphic to $\mathcal{D}^{\pm}(k)$. Then $H(k-2) = 0$ implies that $L_{k}f = 0$ for any $f\in L^{2}(\Gamma \backslash\mathcal{H}, \chi, k)$. This is equivalent to $y^{-k/2}f(z)$ to be a holomorphic modular form in $M_{k}(\Gamma, \chi)$. ◻

Proof. We show first that $V\subset \sum_{\rho\in \widehat{K}} V(\rho)$. For $v\in V$, it is fixed by a compact open subgroup $K_{0}$ of $K$, which can be assumed to be normal. Then $$v\in V^{K_{0}} = \bigoplus_{\rho\in \widehat{\Gamma}} V(\rho) \subseteq \sum_{\rho\in \widehat{K}} V(\rho)$$ where $\Gamma = K/K_{0}$, which is finite.

To show that the sum is direct, let’s assume that it is not, so $\sum_{\rho\in S} c_{\rho}v_{\rho} = 0$ for some finite subset $S\subset \widehat{K}$, $v_{\rho}\in V(\rho)$ and $c_{\rho}\in \mathbb{C}$ that are not all zero. If we put $K_{0} = \cap_{\rho\in S}\ker (\rho)$, then we obtain a contradiction to the directness of the summation for $\Gamma = K/K_{0}$.

For the last statement, $V(\rho)\subset V^{\ker(\rho)}$ implies that $V(\rho)$ is finite dimensional if $\pi$ is admissible since $\ker(\rho)$ is an open subgroup. Conversely, if $\pi$ is not admissible, then $V^{K_{0}}$ is infinite dimensional for some open normal subgroup $K_{0}$ of $K$. From $V^{K_{0}} = \oplus_{\rho\in \widehat{K/K_{0}}} V(\rho)$, since $K/K_{0}$ is a finite group, $V(\rho)$ is infinite dimensional for some $\rho$. ◻

Proof. We will show that $G$-invariance of subspace is equivalent to $\mathcal{H}$-invariance, which proves $1\Leftrightarrow 2$. Clearly, $G$-invariant space is also $\mathcal{H}$-invariant. Conversely, let $W\subset V$ be a $\mathcal{H}$-invariant subspace. Assume that $W$ is not $G$-invariant, so that $\pi(g)w\neq w$ for some $g\in G$ and $w\in W$. Now $w$ is fixed by some neighborhood $N$ of the identity, so let $\phi = \frac{1}{|N|} \mathds{1}_{gN}$ then we have $w = \pi(\phi)w = \phi(g)w$, a contradiction.

$3\Rightarrow 2$ is also simple: assume that $V$ is not simple and let $W \subset V$ be a proper $\mathcal{H}$-submodule. From $V = \cup_{K_{0}} V^{K_{0}}$, we can find $K_{0}$ small enough so that $W^{K_{0}}$ is a nonzero proper subspace of $V^{K_{0}}$.

For $2\Rightarrow3$, let $W_{0}\subset V^{K_{0}}$ be a nonzero proper $\mathcal{H}_{K_{0}}$-submodule. We will show that $\pi(\mathcal{H})W_{0} \cap V^{K_{0}} = W_{0}$, which implies that $\pi(\mathcal{H})W_{0}$ is a nonzero proper $\mathcal{H}$-submodule of $V$. Assume that $w = \sum_{i} \pi(\phi_{i})w_{i} \in \pi(\mathcal{H})W_{0}\cap V^{K_{0}}$, where $w_{i}\in W_{0}$. Since $w_{i}\in V^{K_{0}}$ and $w\in V^{K_{0}}$, we have $\pi(\epsilon_{K_{0}})w_{i} = w_{i}$ and $\pi(\epsilon_{K_{0}})w = w$, which shows that $w = \sum_{i} \pi(\epsilon_{K_{0}} * \phi_{i} * \epsilon_{K_{0}}) w_{i}$. However, $\epsilon_{K_{0}} * \phi_{i} * \epsilon_{K_{0}} \in \mathcal{H}_{K_{0}}$ and since $W_{0}$ is $\mathcal{H}_{K_{0}}$-stable, we get $w\in W_{0}$. ◻

Proof. It is known that for any $k$-algebra $R$, structure of simple $R$-module is completely determined by traces of endomorphisms induced by multiplication of elements in $R$. Hence the assumption implies that $V_{1}^{K_{1}}\simeq V_{2}^{K_{1}}$ as $\mathcal{H}_{K_{1}}$-modules for any open compact subgroup $K_{1}$ of $G$.

Let $K_{0}$ be a small open compact subgroup so that $V_{1}^{K_{0}}$ and $V_{2}^{K_{0}}$ are nonzero. By hypothesis, we have an $\mathcal{H}_{K_{0}}$-module isomorphism $\sigma_{K_{0}}:V_{1}^{K_{0}}\to V_{2}^{K_{0}}$, which is unique up to constant by Schur’s lemma. Then for any open subgroup $K_{1} \subset K_{0}$, we can extend $\sigma_{K_{0}}$ uniquely to a $\mathcal{H}_{K_{1}}$-module isomorphism $\sigma_{K_{1}}:V_{1}^{K_{1}}\to V_{2}^{K_{1}}$. Indeed, the existence is in our hypothesis and from $V_{i}^{K_{0}} = \pi_{i}(\epsilon_{K_{0}}) V_{i}^{K_{1}}$ we have \begin{align*}\sigma_{K_{1}}(V_{1}^{K_{0}}) = \sigma_{K_{1}}(\pi_{1}(\epsilon_{K_{0}})V_{1}^{K_{1}}) = \pi_{2}(\epsilon_{K_{0}}) (\sigma_{K_{1}}(V_{1}^{K_{1}})) = \pi_{2}(\epsilon_{K_{0}}) V_{2}^{K_{1}} = V_{2}^{K_{0}},\end{align*} so $\sigma_{K_{1}}|_{V_{1}^{K_{0}}} : V_{1}^{K_{0}} \to V_{2}^{K_{0}}$ is an $\mathcal{H}_{K_{0}}$-module isomorphism, and uniqueness implies that the restriction of $\sigma_{K_{1}}$ and $\sigma_{K_{0}}$ agrees up to scalar, so we can assume that they coincides on $V_{1}^{K_{0}}$ by normalizing. Now we can repeat this for an open compact basis of identities $\{K_{n}\}_{n\geq 0}$, and we get a map $\sigma:V_{1}\to V_{2}$ which is an $\mathcal{H}$-module isomorphism.

To show that $\sigma$ is an intertwining operator, let $g\in G$ and $v\in V_{1}$. Choose an open compact subgroup $K_{1}$ such that $v\in V^{K_{1}}$, and let $\phi = \frac{1}{|K_{1}|} \mathds{1}_{gK_{1}}$. Then $\pi_{1}(\phi)v = \pi_{1}(g)v$ and $\pi_{2}(\phi)\sigma(v)= \pi_{2}(g)\sigma(v)$, and we get $$\sigma(\pi_{1}(g)v) = \sigma(\pi_{1}(\phi)v) = \pi_{2}(\phi)\sigma(v) = \pi_{2}(g)\sigma(v).$$ This shows that $\sigma$ is an intertwining operator between $V_{1}$ and $V_{2}$. ◻

Proof. For $\phi\in C^{\infty}(G)$, let $\phi', \phi''\in C^{\infty}(G)$ as $\phi'(g) = \phi(g^{-1})$, $\phi''(g) = \phi(\prescript{T}{}{}g^{-1})$. We know that character is conjugation invariant, and it is known that conjugation invariant distribution on $\mathrm{GL}(n, F)$ is also transpose invariant. (This is a nontrivial result proved by Bernstein-Zelevinski. You can found a proof in p. 449 of (bu?).) Hence we have $$\chi_{\pi_{1}}(\phi) = \chi_{\pi}(\phi'') = \chi_{\pi}(\phi') = \chi_{\widehat{\pi}}(\phi)$$ where the last equality follows from the fact that $\pi(\phi)$ and $\widehat{\pi}(\phi')$ are adjoints of each other, so have equal trace.

For 2, the following identity $$\prescript{T}{}{}g^{-1} = \begin{pmatrix} \det(g) & \\ & \det(g) \end{pmatrix}^{-1} w^{-1}gw, \quad w = \begin{pmatrix} & -1 \\ 1 & \end{pmatrix}$$ shows that $\pi(w)$ is an intertwining operator from $(\pi_{1}, V)$ to $(\pi_{2}, V)$. ◻

Proof. $\pi$-invariant subspace is also $\pi_{1}$-invariant. ◻

Proof. It is enough to show that there exists an open neighborhood $N$ of the identity of $\mathrm{GL}(n, \mathbb{C})$ that does not contain any nontrivial open subgroups. Then we can take the compact open subgroup that is contained in $\phi^{-1}(N)$. To show the existence of such $N$, let $\mathfrak{g}= \mathfrak{gl}(n, \mathbb{C})$ be its Lie algebra and let $\exp : \mathfrak{g} \to G'$ be the exponential map. Since $\exp$ is a local homeomorphism, we can find an open neighborhood $U \subseteq \mathfrak{g}$ of the identity such that $\exp:U\to \exp(U)$ is a homeomorphism. Fix an inner product on $\mathfrak{g}$ and we can assume that $U$ is of the form $\{v\in \mathfrak{g}\,:\, |v| <\epsilon$ for some $\epsilon>0$. Let $V = \frac{1}{2} U = \{v\in \mathfrak{g}\,:\, 2v\in U\} = \{v\in \mathfrak{g}\,:\, |v| <\epsilon/2\}$. We will show that $\exp(V)$ contains no nontrivial subgroups. Suppose that $H$ is a nontrivial subgroup contained in $\exp(V)$ and choose $1\neq g\in H$ and $v\in V$ so that $g = \exp(v)$. Since $g^{2}\in H$, $g^{2} = \exp(w)$ for some $w\in V$ and $\exp(2v) = g^{2} = \exp(w)$ implies that $w = 2v$ since $\exp$ is a homeomorphism on $V$. Now iterate this and we have $2^{n}v\in V$ for all $n$, and this implies $v = 0$ since $|2^{n}v| = 2^{n}|v|$. This gives a contradiction since $1\neq g = \exp(v) = \exp(0) = 1$. ◻

Proof. The only nontrivial part for the first exact sequence is the surjectivity of $C_{c}^{\infty}(X) \to C_{c}^{\infty}(C)$. Let $f\in C_{c}^{\infty}(C)$. Since $f$ is locally constant and compactly supported, there exists disjoint open and compact sets $U_{i} \subseteq C$ and $a_{i}\in \mathbb{C}$ such that $f(x) = a_i$ if $x\in U_i$ and $f(x) = 0$ off $\cup U_i$. Let $V_i$ be open and compact subsets of $X$ such that $U_i = V_i \cap C$. By replacing $V_i$ by $V_i \backslash\cup_{j<i} V_j$, we can assume that $V_i$ are disjoint. Then we can extend the function $f$ to $X$ by letting $f(x) = a_i$ if $x\in V_i$ and $f(x) = 0$ off $\cup V_i$. Exactness of the second one follows by dualizing the first one. ◻

Proof. We define another distribution by $f\mapsto T(\xi^{-1}f)$. (Note that $\xi^{-1}f$ is locally constant since $\xi^{-1}$ is locally constant by no small subgroups argument (Proposition \ref{nss}). Replacing $T$ by this, we may assume that $\xi = 1$ so that $\lambda(h)T = T$ for all $h$.

Let $K$ be an open compact subgroup of $G$. If $f\in C_{c}^{\infty}(G)$, let $S(f) = \{h\in G\,:\, \lambda(h)f = f\}$. Then $S(f)$ is a compact open subgroup of $G$ and for any open subgroup $K_0$ of $S(f)$, we have $$f = |K_{0}|\sum_{i=1}^{r} a_{i}\lambda(h_{i})\epsilon_{K_{0}}$$ where $h_{1}, \dots, h_{r}$ be representatives of cosets in $K_{0}\backslash G$ on which $f$ does not vanish and $a_{i} = f(h_{i})$. Then $$T(f) = |K_{0}| \left[ \sum_{i=1}^{r}a_{i}\right] T(\epsilon_{K_{0}}).$$ Apply this for $f = \epsilon_K$ so that $S(f) = K$ and $K_{0}$ can be any open subgroup of $K$. Since $[K:K_{0}] = |K|/|K_{0}|$, we have $T(\epsilon_{K_{0}}) = T(\epsilon_K) = c$. For general $f$, we may assume $K_{0} \subseteq K$ and this implies the desired result. ◻

Proof. By replacing $\Delta$ by $\Delta - \prescript{\iota}{}{}\Delta$, we can assume that $\prescript{\iota}{}\Delta = -\Delta$, too. Now we want to show that a Whittaker distribution satisfying the above condition is zero.

The above conditions (transformations laws) on $\Delta$ can be written in a more simpler way. Let $G$ be a semidirect product of the group $N(F)\times N(F)$ and an order 2 cyclic group generated by $\mathcal{I}$ satisfying $\mathcal{I}^{2} = 1$ and $\mathcal{I}(u_{1}, u_{2})\mathcal{I}^{-1} = (\prescript{\iota}{}{}u_{2}^{-1}, \prescript{\iota}{}{}u_{1}^{-1})$ for $(u_{1}, u_{2})\in N(F)\times N(F)$. Let $\chi$ be a character of $G$ defined as $\chi(u_{1}, u_{2})= \psi_{N}(u_{1})^{-1}\psi_{N}(u_{2})$ and $\chi(\mathcal{I}) = -1$. Let $\sigma$ be the action of $G$ on $\mathrm{GL}(n, F), C_{c}^{\infty}(\mathrm{GL}(n, F))$, and $\mathfrak{D}(\mathrm{GL}(n, F))$ by $\sigma(u_{1}, u_{2}) = \lambda(u_{1})\rho(u_{2}), \sigma(\mathcal{I}) =\iota$. Then the conditions on $\Delta$ can be summarized as $\sigma(g) \Delta = \chi(g)\Delta$.

To show that such distribution is zero, we will use the Bruhat decomposition and the corresponding exact sequence of distributions. We have an exact sequence $$0 \to \mathfrak{D}(B(F)) \to \mathfrak{D}(\mathrm{GL}(2, F)) \to \mathfrak{D}(X) \to 0$$ where $X = B(F)w^{0}B(F)$.

We first show that the image in $\mathfrak{D}(X)$ of $\Delta$ is zero. For a continuous mapping $p:X\to Y$ with $Y = F^{\times} \oplus F^{\times}$ given by $\left(\begin{smallmatrix} a & b \\ c & d \end{smallmatrix}\right) \mapsto (c, (ad-bc)/c)$, the fibers of this map are $\sigma$-invariant and they are the double cosets $$N(F) \begin{pmatrix} & b_{0} \\ c_{0} & \end{pmatrix} N(F)$$ which is homeomorphic to $N(F)\times N(F)$ under the map $(u_{1}, u_{2})\mapsto u_{1} \left(\begin{smallmatrix} & b_{0} \\ c_{0} & \end{smallmatrix}\right) u_{2}^{-1}$. By the theorem of Bernstein-Zelevinski (Proposition \ref{bz}), we only need to show that there are no nonzero distributions $\Delta$ on the single double coset that satisfies $\sigma(g)\Delta = \chi(g)\Delta$. By Proposition \ref{distuniq}, there exists $c\in \mathbb{C}$ such that $$\Delta(f) = c \int\limits_{N(F)\times N(F)} \psi_{N}(u_{1})\psi_{N}(u_{1}) f\left( u_{1} \begin{pmatrix} & b_{0} \\ c_{0} & \end{pmatrix} u_{2}\right) du_{1}du_{2}.$$ This distribution is invariant under $\iota$, since $\iota\left( u_{1} \left(\begin{smallmatrix} & b_{0} \\ c_{0} & \end{smallmatrix}\right) u_{2}\right) = u_{2}\left(\begin{smallmatrix} & b_{0} \\ c_{0} & \end{smallmatrix}\right) u_{1}$. Thus $\prescript{\iota}{}{}\Delta = \Delta = -\prescript{\iota}{}{}\Delta$ and so $\Delta = 0$. By exactness, $\Delta \in \mathfrak{D}(B(F))$. We can use the similar argument to show $\Delta = 0$. Let $Y_{1} =F^{\times} \oplus F^{\times}$ and let $p:B(F) \to Y_{1}, \left(\begin{smallmatrix} a & b \\ & d \end{smallmatrix}\right) \mapsto (a, d)$. Then each fibers are homeomorphic to $N(F)$ via $u\mapsto u\delta = u\left(\begin{smallmatrix} a & \\ & d \end{smallmatrix}\right)$. If we apply the Proposition \ref{distuniq} for left and right translations, we get $$\Delta(\phi) = c_{1}\int\limits_{N(F)} \phi(u\delta)\psi_{N}(u)du = c_{2} \int\limits_{N(F)} \phi(u\delta) \psi_{N}(\delta^{-1}u\delta) du$$ for some $c_{1}, c_{2}\in \mathbb{C}$, where $du$ is the right Haar measure of $N(F)$. If $a\neq d$, then $c_{1} =c_{2} = 0$: otherwise, we may choose $u$ such that $c_{1}\psi_{N}(u) \neq c_{2}\psi_{N}(\delta^{-1}u\delta)$, and then taking a test function $\phi$ that is the characteristic function of a small neighborhood of $u$ gives a contradiction. If $a = d$, then $\prescript{\iota}{}\Delta = \Delta$ so $\Delta = 0$. ◻

proof of Theorem \ref{nonarchmultone} when $n = 2$. The representation $\pi’(g) = \pi(\iota(g^{-1}))$ is isomorphic to $\pi_1$, so to $\hat{\pi}$. Hence we have a pairing $V \times V \to \mathbb{C}$ s.t. $\langle \pi(g)v,w \rangle = \langle v,\pi(\iota(g))w\rangle$. By Riesz representation theorem, any linear functional $\Lambda$ corresponds to a vector $[\Lambda]$ by $\langle v, [\Lambda]\rangle = \Lambda(v)$. We can also define another linear functional $\Lambda * \phi$ for any $\phi \in \mathcal{H}$ by $$(\Lambda * \phi)(\xi) = \Lambda(\pi(\phi)\xi) = \int_{G} \Lambda(\pi(g)\xi)\phi(g)dg$$ which satisfies the associativity $\Lambda * (\phi_1 * \phi_2) = (\Lambda * \phi_1) * \phi_2$. It satisfies following transformation laws: \begin{align*}\pi(g) [\Lambda*\phi] &= [\Lambda * \rho(\iota(g^{-1}))\phi] \\ [L* \phi] &= \pi(\prescript{\iota}{}{}\phi)[L]\\ [\Lambda * \lambda(u)\phi] &= \psi_N(u)[\Lambda * \phi]\end{align*} (Second one holds for smooth $L$, and the third one holds for Whittaker functionals.) Now if $\Lambda_1,\Lambda_2$ are Whittaker functionals, define a distribution $\Delta(\phi) = \Lambda_2([\Lambda_1 * \phi])$. The above transformation properties imply that this is a Whittaker distribution, so it is invariant under the involution by the previous theorem. Using that, we can show $\Lambda_1 * \phi = 0 \Rightarrow \Lambda_2 * \phi = 0$. One can show that for any given nonzero linear functional $\Lambda$ on $V$, any vector in V has a form of $[\Lambda * \phi]$ for some $\phi\in \mathcal{H}$. Then we can define a map $T:[\Lambda_1 * \phi] \mapsto [\Lambda_2 * \phi]$, which is an intertwining map by the above transformation law. By Schur’s lemma, $T = cI$ for some $c$, and this gives $\Lambda_2 = c\Lambda_1$. ◻

Proof. Let $0\to V'\to V\to V'' \to 0$ be a short exact sequence of $B(F)$-modules. Then we can prove that the induced sequence $0\to V_{N}' \to V_N \to V_{N}'' \to 0$ is also exact. Here we use the following characterization: $x\in V_N$ iff $$\int_{\mathfrak{p}^{-n}} \pi\begin{pmatrix} 1 & x \\ & 1 \end{pmatrix} v dx = 0$$ for sufficiently large $n$. Now we get the result by the snake lemma. Proof for $J_{\psi}$ is similar. ◻

Proof. Assume that $(\pi, V)$ has no nonzero Whittaker functional. Fix an additive char $\psi$ of $F$ and corresponding self-dual Haar measure. Let $\psi_a:F\to \mathbb{C}^{\times}$ be a character $\psi_a(x)=\psi(ax)$. Then the stalk of the above sheaf is given by $$\mathcal{S}(V)_{a} \simeq \begin{cases} J(V) & a = 0 \\ J_{\psi_{a}}(V) \simeq J_{\psi}(V) & a\neq 0 \end{cases}$$ so $\mathcal{S}(V)_a = 0$ for $a\neq 0$ and $\mathcal{S}(V)$ is a skyscraper sheaf at $a=0$. Then $V\to S(V)_0=J(V)$ is an isomorphism, so that $V_N=0$ and $N(F)$ acts trivially. Then all conjugates of $N(F)$ also acts trivially, so does $\mathrm{SL}(2, F)$ (they generate $\mathrm{SL}(2,F)$) and factors through determinant map. ◻

Proof. For $\Phi\in \mathrm{Hom}_G(\sigma, \mathrm{Ind}_{H}^{G}\pi)$, define $\phi\in \mathrm{Hom}_H(\sigma|_{H}, \pi\otimes (\delta_{G}^{-1}\delta_{H})^{1/2})$ by $\phi(w) = \Phi(w)(1)$. Conversely, if $\phi\in \mathrm{Hom}_H(\sigma|_{H}, \pi\otimes (\delta_{G}^{-1}\delta_{H})^{1/2})$ is given, we can define $\Phi\in \mathrm{Hom}_G(\sigma, \mathrm{Ind}_{H}^{G}\pi)$ as $\Phi(w)(g) = \phi(\sigma(g)w)$. It is easy to check that these maps give isomorphisms between two spaces. ◻

Proof. Use induction on $n$. One can find $k_1\in K$ such that $$gk_{1} =\begin{pmatrix} g_{n-1} & * \\ 0 & x_{n} \end{pmatrix}$$ for some $g_{n-1}\in \mathrm{GL}(n-1, F)$. By induction hypothesis, there exists $k'\in \mathrm{GL}(n-1, \mathcal{O}_F)$ such that $g_{n-1}k'$ is upper triangular. Then $k = k_1 \left(\begin{smallmatrix} k' & \\ & 1 \end{smallmatrix}\right)$ makes $gk \in B(F)$. Then the map $K \to B(F)\backslash G$ is continuous and surjective, so the coset $B(F)\backslash G$ is compact. ◻

Proof. Define $P:C^{\infty}_{c}(\mathrm{GL}(2,F))\to V$ by convolutioning with $\delta^{1/2}\chi$ over $B(F)$, where $\delta = \delta_{B(F)}$, $$(P\phi)(g) = \int\displaylimits_{B(F)} \phi(b^{-1}g) (\delta^{1/2}\chi)(b)db.$$ (Here $db = d_{L}b$.) One can show that $P$ is surjective by choosing $\phi = \frac{1}{|K\cap B(F)|}\mathds{1}_{K}f$ for $f\in V$. Also, it satisfies $P(\lambda(b)^{-1}\phi) = (\delta^{-1/2}\chi)(b)P(\phi)$ and $P(\rho(g)\phi) = \pi(g)P(\phi)$ for all $b\in B(F)$ and $g\in \mathrm{GL}(2, F)$.

Now let $\Lambda: V\to \mathbb{C}$ be a Whittaker functional. Define $\Delta\in \mathfrak{D}(\mathrm{GL}(2, F))$ as $\Delta(\phi) = \Lambda(P\phi)$. Then it satisfies $\lambda(b) \Delta = (\delta^{-1/2}\chi)(b)\Delta$ and $\rho(n)\Delta = \psi_{N}(n)^{-1}\Delta$ for all $b\in B(F)$ and $n\in N(F)$. Now we use Bruhat decomposition again: we have an exact sequence $$0\to \mathfrak{D}(B(F)) \to \mathfrak{D}(\mathrm{GL}(2, F)) \to \mathfrak{D}(X) \to 0$$ where $X = \mathrm{GL}(2, F) - B(F) = B(F)w_{0}N(F)$ where $w_{0} = \left(\begin{smallmatrix} & -1 \\ 1 & \end{smallmatrix}\right)$. Let $\Delta_1\in \mathfrak{D}(X)$ be a distribution satisfies the above transformation laws. By the Proposition \ref{distuniq}, there exists $c\in \mathbb{C}$ such that $$\Delta_{1}(\phi) = c\int\displaylimits_{B(F)}\int\displaylimits_{N(F)} \phi(bw_{0}n^{-1})\psi_{N}(n)(\delta^{1/2}\chi^{-1})(b)db\,dn.$$ Also, let $\Delta_2\in \mathfrak{D}(B(F))$ be a distribution satisfies the above transformation laws. By the Proposition \ref{distuniq} again, there exists $c\in \mathbb{C}$ such that $$\Delta_2(\phi) = c\int\displaylimits_{B(F)} \phi(b) (\delta^{1/2}\chi^{-1})(b)db$$ for all $\phi\in C_{c}^{\infty}(B(F))$. Then $\rho(n)\Delta_2 = \Delta_2$ for $n\in N(F)$, so together with $\rho(n)\Delta_2 = \psi_{N}(n)^{-1}\Delta_2$ we get $\Delta_2 =0$. By combining these two results, we can show that the space of $\Delta\in \mathfrak{D}(\mathrm{GL}(2, F))$ satisfying the above equations is one dimensional. ◻

Proof. Let $(\pi, V) = \mathcal{B}(\chi_1, \chi_2)$ and $(\pi', V') = \mathcal{B}(\chi_1^{-1}, \chi_{2}^{-1})$. Then we can define a non-degenerate pairing $\langle\,,\,\rangle : V\times V' \to \mathbb{C}$ by $$\langle f, f'\rangle = \int\displaylimits_{K} f(k)f'(k)dk$$ which is $G$-invariant. ◻

Proof. For a fixed vector $f$, $\pi$ acts as a character, which factors through commutator subgroup $\mathrm{SL}(2,F)$, hence $\pi(g) = (\rho\circ\det)(g)$ for some quasi-character $\rho$. Then $f(bg) = (\delta^{1/2} \chi)(b) \rho(\det(g)) f(1)$, and taking $b = g^{-1} =\left(\begin{smallmatrix} y & \\ & y^{-1} \end{smallmatrix}\right)$ gives the result. Second one follows from taking dual of first one. ◻

Proof. We use exactness of twisted Jacquet module and its relation to Whittaker models. Assume that $V$ is not irreducible, so it has a nontrivial proper invariant subspace $0\subsetneq V'\subsetneq V$. Let $V'' = V/V'$ and let $\pi', \pi''$ be the corresponding representations. Then we get an exact sequence $$0\to J_{\psi}(V') \to J_{\psi}(V) \to J_{\psi}(V'') \to 0.$$ Since $\dim J_\psi(V) \leq 1$, at least one of $J_\psi(V')$ or $J_{\psi}(V'')$ is zero. If $J_{\psi}(V') = 0$, then $\pi'$ factors through $\det : \mathrm{GL}(2, F) \to F^{\times}$ by Theorem \ref{dim1}. One can prove that admissible representation of $F^{\times}$ contains a 1-dimensional invariant subspace, so we get $\chi_1 \chi_2^{-1} = |\cdot|^{-1}$ by the previous lemma. In this case, the function $f(g) = \chi(\det(g))$ spans an invariant 1-dimensional subspace when $\chi_{1}(y) = \chi(y)|y|^{-1/2}$ and $\chi_{2}(y) = \chi(y)|y|^{1/2}$. We can prove another case $J_\psi(V'') = 0$ by dualizing. ◻

Proof. By Frobenius reciprocity, intertwining map corresponds to a linear functional $\Lambda:V\to \mathbb{C}$ with $B(F)$-module structure on $\mathbb{C}$ by means of the quasi-character $\delta^{1/2}\mu$. Then $\Delta = \Lambda \circ P$ (here $P:C_{c}^{\infty}(\mathrm{GL}(2, F))\to V$ is the convolutioning map we defined in Theorem \ref{psrepwh}) is a nonzero distribution and satisfies $\lambda(b)\Delta = (\delta^{-1/2}\chi)(b)\Delta$ and $\rho(b)\Delta = (\delta^{-1/2}\mu^{-1})(b)\Delta$ for $b\in B(F)$. From the exact sequence of distribution, there exists a nonzero distribution that satisfies same equations in either $\mathfrak{D}(B(F))$ or $\mathfrak{D}(\mathrm{GL}(2, F) - B(F))$.

First, assume that $\Delta\in \mathfrak{D}(\mathrm{GL}(2, F) - B(F))$ is a such nonzero distribution. By Proposition \ref{distuniq}, we have $$\Delta(\phi) = \int\displaylimits_{B(F)}\int\displaylimits_{N(F)} \phi(bw_{0}n^{-1}) (\delta^{1/2}\chi^{-1})(b) db\,dn$$ after adjusting by a nonzero constant. If we apply $\rho(t)$ action and using a change of variable, we have \begin{align*}(\delta^{-1/2}\mu^{-1})(t)\Delta(\phi) &= (\rho(t)\Delta)(\phi) \\ &=\int\displaylimits_{N(F)}\int\displaylimits_{B(F)} \phi(bw_{0}n^{-1}t^{-1})(\delta^{1/2}\chi^{-1})(b)db\,dn \\ &=\delta(t)^{-1}(\delta^{1/2}\chi^{-1})(w_{0}tw_{0}^{-1})\Delta(\phi).\end{align*} where the last equality follows from the change of variables $n\mapsto t^{-1}nt, b\mapsto bw_{0}tw_{0}^{-1}$. From $\delta(t) = \delta(w_{0}tw_{0}^{-1})^{-1}$, we have $\mu(t) = \chi(w_{0}tw_{0}^{-1})$ and so $\chi_1 = \mu_2$ and $\chi_2 = \mu_1$.

Another case is similar. By using the integral form of the distribution with change of variables, we can show that for $b\in B(F)$ we have $(\delta^{-1/2}\mu^{-1})(b)\Delta = \rho(b)\Delta = (\delta^{-1/2}\chi^{-1})(b)\Delta$, so $\chi_1 = \mu_1$ and $\chi_2 = \mu_2$. ◻

Proof. We have \begin{align*}f\left(\begin{pmatrix} & -1 \\ 1 & \end{pmatrix}\begin{pmatrix} 1 & x \\ & 1 \end{pmatrix}g\right) &= f\left( \begin{pmatrix} x^{-1} & -1 \\ & x \end{pmatrix} \begin{pmatrix} 1 & \\ x^{-1} & 1 \end{pmatrix} g\right) \\ &= |x|^{-1}(\chi_{1}^{-1}\chi_{2})(x) f\left( \begin{pmatrix} 1 & \\ x^{-1} & 1 \end{pmatrix}g\right).\end{align*} Since $f$ is smooth, there exists $N$ such that if $|x|>q^N$, then $$f\left( \begin{pmatrix} 1 & \\ x^{-1} & 1 \end{pmatrix}g\right) = f(g).$$ Hence the absolute convergence of the integral is equivalent to the convergence of $$\int\displaylimits_{|x|>q^N} |x|^{-1}|(\chi_{1}^{-1}\chi_{2})(x)| dx = \int\displaylimits_{|x|>q^N} |x|^{-s_{1} + s_{2} - 1} dx$$ and this converges if $\Re(s_1 - s_2) >0$. ◻

Proof. Fix $\xi_1, \xi_2, f_{0}\in V_{0}$ and $g$. Since $f$ is smooth, there exists $N$ such that \begin{align*}Mf_{s_{1}, s_{2}}(g) &= \int\displaylimits_{|x|\leq q^{N}} f_{s_{1}, s_{2}}\left( \begin{pmatrix} & -1 \\ 1 & \end{pmatrix} \begin{pmatrix} 1 & x \\ & 1 \end{pmatrix} g\right) dx \\ &+ \int\displaylimits_{|x|\geq q^{N+1}} |x|^{-s_{1} + s_{2} -1} (\xi_{1}^{-1}\xi_{2})(x) dx \cdot f_{s_{1},s_{2}}(g). \end{align*} The first integral converges absolutely (since the domain is compact), so the analytic continuation is clear. For the second integral, if $\xi_{1}\xi_{2}^{-1}$ is ramified then $$\int\displaylimits_{|x| = q^{m}} (\xi_{1}^{-1}\xi_{2})(x)dx = 0$$ for all $m\in \mathbb{Z}$ (consider the change of variable $x\mapsto \alpha x$ with $\alpha\in \mathcal{O}_{F}^{\times}$, $(\xi_{1}^{-1}\xi_{2})(\alpha)\neq 1$) and so the integral vanishes. If $\xi_{1}\xi_{2}^{-1}$ is unramified then there exists $\alpha\in \mathbb{C}$ such that $(\xi_{1}\xi_{2}^{-1})(y) = \alpha^{v(y)}$ where $v:F^{\times} \to \mathbb{Z}$ is the valuation map. Then the integral became \begin{align*}&\sum_{m = N+1}^{\infty}\int\displaylimits_{|x|= q^{m}} |x|^{-s_{1} + s_{2}} (\xi_{1}^{-1}\xi_{2})(x) \frac{dx}{|x|} f_{s_{1}, s_{2}}(g) \\ &= |\mathcal{O}_{F}^{\times}| f_{s_{1}, s_{2}}(g) \sum_{m \geq N+1} (\alpha q^{-s_{1} + s_{2}})^{m}. \end{align*} The latter sum equals constant times $(\alpha q^{-s_{1} + s_{2}})^{N+1} (1-\alpha q^{-s_{1}+s_{2}})^{-1}$ for $\Re(s_{1} - s_{2}) >0$, and it has analytic continuation for all $s_{1}, s_{2}$, except where $\alpha q^{-s_{1} +s_{2}} = 1 \Leftrightarrow \chi_{1} = \chi_{2}$. This also proves that the analytically continued integral remains an intertwining operator and nonzero. ◻

Proof. The proof is based on the uniqueness of Whittaker models. We have a Whittaker functional $\Lambda:\mathcal{B}(\chi_1, \chi_2)\to \mathbb{C}$ defined by $$\Lambda (f) = \int\displaylimits_{F} f\left( w_{0}\begin{pmatrix} 1 & x \\ & 1 \end{pmatrix}\right)\psi(-x) dx.$$ This is absolutely convergent if $\Re(s_1 - s_2) >0$, and we also have an analytic continuation for all $s_1, s_2$. Similarly, we have a Whittaker functional $\Lambda' : \mathcal{B}(\chi_2, \chi_1)\to \mathbb{C}$. By uniqueness, there exists $\lambda\in \mathbb{C}$ such that $(\Lambda' \circ M)(f) = \lambda\Lambda(f)$. We can compute the constant $\lambda$ by inserting suitable $f$. This gives us $\lambda = \xi_{1}\xi_{2}^{-1}(-1) \gamma(1-s_1 + s_2, \xi_1^{-1}\xi_2, \psi)$, which implies the desired result. For detail, see Proposition 4.5.9 and 4.5.10 in (bu?). ◻

Proof. One can always find 1-dimensional invariant subspace, so there exists a nonzero vector fixed by the action. Hence there exists a character $\xi:F^{\times} \to \mathbb{C}^{\times}$ such that $\rho(t)x = \xi(t)x$ for all $t\in F^{\times}$. Since $V/\mathbb{C}x$ is 1-dimensional again, $F^{\times}$ acts by quasi-character $\xi'$ on this space. If we choose $y\in V - \mathbb{C}x$, then we have $$\rho(t)y = \xi'(t)y + \lambda(t)x$$ for some $\lambda:F^{\times} \to \mathbb{C}$. From $\rho(tu) = \rho(t)\rho(u)$, we get $$\lambda(tu) = \xi'(u)\lambda(t) + \lambda(u)\xi(t).$$ When $\xi\neq \xi'$, one can show that $\lambda(t) = C(\xi(t) - \xi'(t))$ for some $C\in \mathbb{C}$, and then $\rho(t)$ has the above form of diagonal matrix with respect to the basis $\{x, y - Cx\}$. If $\xi = \xi'$, $t\mapsto \lambda(t)/\xi(t)$ is a homomorphism from $F^{\times}$ to $\mathbb{C}$, and $\mathcal{O}_F^{\times}$ lies in the kernel since the image is a compact subgroup of $\mathbb{C}$, which is trivial. Hence $\lambda(t) = c\xi(t)v(t)$ for some $c\in \mathbb{C}$, and $\rho$ is isomorphic to the first one if $c =0$, or the second one if $c\neq 0$. ◻

Proof. First, one can show that $\dim J(\mathcal{B}(\chi_1, \chi_2)) = 2$ for any $\chi_1, \chi_2$. To show this, we can construct two explicit linearly independent linear functionals on $J(V)$, which shows $\dim J(V) \geq 2$. The opposite direction uses the Bruhat decomposition, the exact sequence of distributions, and the Proposition \ref{distuniq}.

Now consider $T_{1}(F) = \left\{ \left(\begin{smallmatrix} a & \\ & 1 \end{smallmatrix}\right)\,:\, a\in F^{\times}\right\}$. Since $T_1(F)\simeq F^{\times}$, the action is isomorphic to one of the representations in the previous proposition. Since $T(F) = T_{1}(F)Z(F)$ and $Z(F)$ acts as a scalar, $(T(F), J(V))$ is isomorphic to one of the following representations $$t\mapsto \begin{pmatrix} \delta^{1/2}\xi(t) & \\ & \delta^{1/2}\xi'(t) \end{pmatrix}\quad\text{or}\quad \delta^{1/2}\xi(t)\begin{pmatrix} 1 & v(t_1/t_2) \\ & 1 \end{pmatrix}.$$ To distinguish two cases, we use \begin{align*}\mathrm{Hom}_{T(F)}(J(V), \delta^{1/2}\eta) \simeq \mathrm{Hom}_{B(F)}(V, \delta^{1/2}\eta) \simeq \mathrm{Hom}_{\mathrm{GL}(2, F)}(V, \mathcal{B}(\eta_1, \eta_2))\end{align*} where $\eta_1, \eta_2$ are quasi-characters of $F^{\times}$ and $\eta\left(\begin{smallmatrix} t_1 & \\ & t_2 \end{smallmatrix}\right) = \eta_1(t_1)\eta_2(t_2)$ is a quasi-character of $T(F)$, trivially extended to $B(F)$. If $\chi_1 \neq \chi_2$, then the $\mathrm{Hom}$ is nonzero iff $\eta = \chi$ or $\eta = \chi'$, and this is the case when $(\pi_N, J(V))$ has a form of diagonal matrices with $\xi = \chi$ and $\xi' = \chi'$. If $\chi_1 = \chi_2$, then the $\mathrm{Hom}$ is nonzero iff $\eta_1 =\eta_2 = \chi_1 = \chi_2$, in which chase it is 1-dimensional by Schur’s lemma because $\mathcal{B}(\chi_1, \chi_2)$ is irreducible. This is possible only if $(\pi_N, J(V))$ is of the second form with $\xi = \chi$. ◻

Proof. The proof uses Mackey’s theory, which gives an explicit description of decomposition of the space $\mathrm{Hom}_{G}(V_{1}, V_{2})$ where both $V_{1}, V_{2}$ are induced representations form some closed subgroups of $G$. For the detailed proof, see the Theorem 4.8.1 of (bu?). ◻

Proof. For supercuspidality, we can show that restriction of $\omega_{\xi}$ to $B_{1}(F)$ is isomorphic to $\mathrm{cInd}_{N(F)}^{B_{1}(F)}(\psi_{N})$, which is a $B_{1}(F)$-representation on the space $C_{c}^{\infty}(F^{\times})$. Since $V_N = C_{c}^{\infty}(F^{\times})$ and it is irreducible, we get $V = V_N$ and $\omega_{\xi}$ is a supercuspidal representation. For details and proof of smoothness, admissibility and irreducibility, see 542p of (bu?). ◻

Proof. By Theorem 3.2, it is enough to show that $\pi_{1}(g) = \pi(\prescript{T}{}{g}^{-1})$ is spherical. Since $K$ is invariant under transpose, the spherical vector for $\pi$ is also spherical vector for $\pi_1$. ◻

Proof. We use $p$-adic version of Cartan decomposition theorem: a complete set of double coset representatives for $K\backslash\mathrm{GL}(2, F)/K$ consists of diagonal matrices $$\begin{pmatrix} \varpi^{n_1} & \\ & \varpi^{n_2} \end{pmatrix}$$ where $n_1 \geq n_2$ are integers. Proof is almost same as archimedean case. Now define $\iota:\mathcal{H}_K\to \mathcal{H}_K$ by $\prescript{\iota}{}{\phi}(g) = \phi(\prescript{T}{}{g})$. Then $\prescript{\iota}{}{(\phi_1 * \phi_2)} = \prescript{\iota}{}{\phi_2} * \prescript{\iota}{}{\phi_1}$ holds by direct computation. By the Cartan decomposition, double cosets are invariant under transpose and so $\iota$ is the identity map. This implies $\phi_1 * \phi_2 = \phi_2 * \phi_1$. ◻

Proof. Assume that $V^{K} \neq 0$. By Proposition \ref{simple}, $V^{K}$ is a finite dimensionalsimple $\mathcal{H}_K$-module, so is 1-dimensional since $\mathcal{H}_K$ is commutative. The second assertion follows from the first assertion, since such $L$ would be a $K$-fixed vector in the contragredient representation. ◻

Proof. By Proposition \ref{simple}, it is sufficient to show that $V_{1}^{K_1}\simeq V_{2}^{K_1}$ as $\mathcal{H}_{K_{1}}$-modules for any open subgroup $K_1\subseteq K$. It is known that such $\mathcal{H}_{K_{1}}$-module structures are determined by matrix coefficients, i.e. a function $c:\mathcal{H}_{K_1} \to \mathbb{C}$ of the form $c_i(\phi) = L_i(\pi(\phi)x)$ for a linear functional $L:V_i\to \mathbb{C}$ and $x\in V_i$. So it is enough to show that $$\langle \pi_1(\phi)v_1, \widehat{v_1}\rangle = \langle \pi_2(\phi)v_2, \widehat{v_2}\rangle$$ for any $\phi\in \mathcal{H}$, where $v_i, \widehat{v_i}$ are normalized spherical vectors of $V_i, \widehat{V_i}$ so that $\langle v_i, \widehat{v_i}\rangle = 1$ for $i= 1,2$. If we define $P: \mathcal{H}\to \mathcal{H}$ as $P(\phi) = \epsilon_K * \phi * \epsilon_K$, then we have $P^{2} = P$, $\mathrm{Im}(P) = \mathcal{H}_K$, and $\mathcal{H}= \mathrm{Im}(P) \oplus \ker(P) = \mathcal{H}_K \oplus \ker(P)$. So it is enough to show for $\phi\in \mathcal{H}_K$ and $\phi\in \ker(P)$ separately. We can easily check that for $\phi \in \mathcal{H}_K$, both sides are same as $\xi(\phi)$, and for $\phi\in \ker(P)$, both sides vanishes (here we use $\pi_i(\epsilon_K) v_i = v_i$ and $\widehat{\pi_i}(\epsilon_K) = \widehat{v_i}$. Now restrict the equation for $\phi\in \mathcal{H}_{K_1}$ for $K_1\subseteq K$ and we get $V_1^{K_1}\simeq V_2^{K_1}$ as $\mathcal{H}_{K_1}$-modules. ◻

Proof. Since $T(\mathfrak{p})*T(\mathfrak{p}^k), T(\mathfrak{p}^{k+1}), R(\mathfrak{p})*T(\mathfrak{p}^{k-1})$ are all supported on the double cosets whose determinants generate the ideal $\mathfrak{p}^{k+1}$, so it is sufficient to verify it for the matrices $$\begin{pmatrix} \varpi^{k+1-r} & \\ & \varpi^{r} \end{pmatrix}, \quad r\in \mathbb{Z}$$ with $0\leq r\leq k+1$. Using $$K \begin{pmatrix} \varpi & \\ & 1 \end{pmatrix}K = \begin{pmatrix} 1 & \\ & \varpi \end{pmatrix}K \cup \bigcup_{b\,(\mathrm{mod}\,\mathfrak{p})} \begin{pmatrix} \varpi & b \\ & 1 \end{pmatrix} K$$ we can prove $$(T(\mathfrak{p}) * T(\mathfrak{p}^{k}))(g) = T(\mathfrak{p}^{k}) \left( \begin{pmatrix} 1 & \\ & \varpi \end{pmatrix}^{-1}g\right) + \sum_{b\,(\mathrm{mod}\,\mathfrak{p})} T(\mathfrak{p}^{k}) \left( \begin{pmatrix} \varpi & b \\ & 1 \end{pmatrix}^{-1} g\right)$$ and we get the result by comparing both sides for the above matrices. ◻

Proof. By Cartan decomposition, a basis of $\mathcal{H}_K$ are characteristic functions of the double cosets $$K\begin{pmatrix} \varpi^n & \\ & \varpi^m \end{pmatrix} K, \quad n\geq m.$$ which equals $R(\mathfrak{p})^m$ times the characteristic function of $$K\begin{pmatrix} \varpi^{n-m} & \\ & 1 \end{pmatrix} K.$$ or $T(\mathfrak{p}^{n-m}) - R(\mathfrak{p})*T(\mathfrak{p}^{n-m-2})$. Since $T(\mathfrak{p}^{k})$ can be generated by $T(\mathfrak{p})$ and $R(\mathfrak{p})$, we are done. ◻

Proof. Let $\chi\left(\begin{smallmatrix} b_{1} & * \\ & b_{2} \end{smallmatrix}\right) = \chi_{1}(b_{1})\chi_{2}(b_{2})$ be a quasicharacter of $B(F)$. Let $\phi_{K, \chi}:\mathrm{GL}(2, F)\to \mathbb{C}$ be a function defined as $\phi_{K, \chi}(bk) = (\delta^{1/2}\chi)(b)$. (Recall that any elements in $\mathrm{GL}(2, F)$ can be written as a form of $bk$ for $b\in B(F)$ and $k\in K$ by Iwasawa decomposition.) Well-definedness follows from unramifiedness of $\chi_1, \chi_2$. Also, it is clear that $\phi_{K, \chi}$ is a spherical vector. ◻

Proof. We use the previous decomposition of $K\left(\begin{smallmatrix} \varpi & \\ & 1 \end{smallmatrix}\right)K$. Since $\phi_K(1) = 1$, \begin{align*}\lambda &= (T(\mathfrak{p})\phi_K)(1) = \int\displaylimits_{K\left(\begin{smallmatrix} \varpi & \\ & 1 \end{smallmatrix}\right)K} \phi_{K}(g)dg \\ &= \sum_{\gamma\in K\left(\begin{smallmatrix} \varpi & \\ & 1 \end{smallmatrix}\right)K/K} \int_K \phi_K(\gamma k)dk \\ &= (\delta^{1/2}\chi)\begin{pmatrix} 1 & \\ & \varpi \end{pmatrix} + q(\delta^{1/2}\chi)\begin{pmatrix} \varpi & \\ & 1 \end{pmatrix} = q^{1/2}(\alpha_1 + \alpha_2). \end{align*} For $R(\mathfrak{p})$, it is much easier: $$\mu = (R(\mathfrak{p})\phi_K)(1) = \int\displaylimits_{K\left(\begin{smallmatrix} \varpi & \\ & \varpi \end{smallmatrix}\right)K} \phi_K(g)dg = (\delta^{1/2}\chi)\begin{pmatrix} \varpi & \\ & \varpi \end{pmatrix} = \alpha_1 \alpha_2.$$ ◻

Proof. Let $\xi$ be the character of $\mathcal{H}_K$ associated with the spherical representation $(\pi, V)$ and let $\lambda, \mu$ be the eigenvalues of $T(\mathfrak{p})$ and $R(\mathfrak{p})$. Since $R(\mathfrak{p})$ is invertible, $\mu\neq 0$. Let $\alpha_1, \alpha_2$ be the roots of the quadratic polynomial $X^{2} - q^{-1/2}\lambda X + \mu = 0$, and let $\chi_1,\chi_2$ be the unramified quasicharacters of $F^{\times}$ with $\chi_i(\varpi) = \alpha_i$. Then $T(\mathfrak{p})$ and $R(\mathfrak{p})$ have the same eigenvalues $\lambda$ and $\mu$ on $V^{K}$, so the character $\xi$ and the character associated with $\mathcal{B}(\chi_1, \chi_2)$ coincides. So if $\mathcal{B}(\chi_1, \chi_2)$ is irreducible, $(\pi, V)\simeq \mathcal{B}(\chi_1, \chi_2)$ by the Theorem \ref{sphchar}. If not, we may assume $\alpha_1 \alpha_2^{-1} = q$ so that $\mathcal{B}(\chi_1, \chi_2)$ has a 1-dimensional subspace. By Theorem \ref{sphchar} again, $(\pi, V)$ is isomorphic to this 1-dimensional subrepresentation, and we can check that $\pi(g) = \chi(\det(g))$ where $\chi(y) = |y| \chi_1(y)$. ◻

Proof. It is clear that $M\phi_{K, \chi}$ is a spherical vector in $V'$, so $M\phi_{K, \chi} = c\phi_{K, \chi'}$ for some constant $c = M\phi_{K, \chi}(1) \in \mathbb{C}$. For the computation of $c$, we have \begin{align*}\int\displaylimits_{F} \phi_{K, \chi}\left(\begin{pmatrix} & -1 \\ 1 & \end{pmatrix} \begin{pmatrix} 1 & x \\ & 1 \end{pmatrix}\right) dx &= 1 + \sum_{m\geq 1} |\mathfrak{p}^{-m} - \mathfrak{p}^{-(m-1)}| q^{-m}\alpha_{1}^{m}\alpha_{2}^{-m} \\ &= 1 + \sum_{m\geq 1}(1-q^{-1})(\alpha_{1}\alpha_{1}^{-1})^{m} \\ &= \frac{1-q^{-1}\alpha_{1}\alpha_{2}^{-1}}{1- \alpha_{1}\alpha_{2}^{-1}}.\end{align*} ◻

Proof. For $m<0$, we have $$W_{0}\left( a_m \begin{pmatrix} 1 & x \\ & 1 \end{pmatrix} \right) = W_{0}\left( \begin{pmatrix} 1 & \varpi^{m}x \\ & 1 \end{pmatrix}a_m \right) = \psi(\varpi^{m}x)W_{0}(a_{m})$$ and by choosing $x\in \mathcal{O}_F$ with $\psi(\varpi^{m}x)\neq 1$, we get $W_{0}(a_{m}) = 0$. For $m\geq 0$, we use a special basis $\{\phi_0, \phi_1\}$ of $V^{K_{0}(\mathfrak{p})} \simeq J(V)$, which is called Casselman basis. These are vectors such that $L_{0}(\phi_{0}) = L_{1}(\phi_{1}) = 1$ and $L_{1}(\phi_{0}) = L_{0}(\phi_{1}) = 0$, where $L_{0}, L_{1}$ are linear functionals on $V$ defined as $L_{1}(\phi) = \phi(1)$ and $L_{0}(\phi) = (M\phi)(1)$, which can be regarded as a functional on $J(V)$ or $V^{K_{0}(\mathfrak{p})}$. One can check that these are nonzero and linearly independent, so form a basis of $J(V)^{*} \simeq (V^{K_{0}(\mathfrak{p})})^{*}$. Using this, one can prove that $$W_{0}(a_{m}) = C q^{-m/2}\alpha_{1}^{m} + M\phi_{K, \chi}(1) q^{-m/2}\alpha_2^m,$$ for some $C\in \mathbb{C}$. Also, $(1-q^{-1}\alpha_{1}\alpha_{2}^{-1})^{-1}W_{0}(g)$ is invariant under the interchange of $\alpha_1$ and $\alpha_2$ since it is a normalized spherical vector and $\mathcal{B}(\chi_1, \chi_2)\simeq \mathcal{B}(\chi_2, \chi_1)$. Using this, we can compute $C$ and we obtain the formula. ◻

Proof. For $f_1, f_2\in V$, $$\langle f_1, f_2\rangle = \int\displaylimits_{K} f_1(k)\overline{f_2(k)} dk$$ defines a positive-definite $\mathrm{GL}(2, F)$-invariant Hermitian pairing. ◻

Proof. If $\langle \,,\,\rangle$ is the paring, then $(f_1, f_2) := \langle f_1, \overline{f_2}\rangle$ is a nondegenerate $\mathrm{GL}(2, F)$-invariant bilinear pairing $$\mathcal{B}(\chi_1, \chi_2) \times \mathcal{B}(\overline{\chi}_1, \overline{\chi}_{2})\to \mathbb{C},$$ and so $\mathcal{B}(\overline{\chi}_{1}, \overline{\chi}_2) \simeq \widehat{\mathcal{B}(\chi_{1}, \chi_2)} \simeq \mathcal{B}(\chi_1^{-1}, \chi_2^{-1})$. ◻

Proof. We may assume $s<0$. Since $V = \mathcal{B}(\chi_s, \chi_s^{-1})$ is irreducible, there can be at most one nondegenerate $\mathrm{GL}(2, F)$-invariant sesquilinear pairing on $V$ (up to a constant multiple). If we put $M_{s}: \mathcal{B}(\chi_s, \chi_s^{-1})\to \mathcal{B}(\chi_s^{-1}, \chi_s)$, then we get a nondegenerate sesquilinear pairing $$\langle f_1, f_2\rangle = \int\displaylimits_{K} (M_s f_1)(k) \overline{f_{2}(k)} dk.$$ Define an Iwahori fixed vector $$f_{0}(g) = \begin{cases} \delta^{s+1/2}(b) & g = bk, b\in B(F), k_{0} \in K_{0}(\mathfrak{p}) \\ 0 & g \in B(F)w_{0}K_{0}(\mathfrak{p}) \end{cases}$$ where $K_{0}(\mathfrak{p}) = \{ \left(\begin{smallmatrix} a & b \\ c & d \end{smallmatrix}\right)\in \mathrm{GL}(2 ,\mathcal{O}_{K})\,:\, c\in \mathfrak{p}\}$. Then we get $$\langle f_0, f_0\rangle = \frac{1-q^{-1}}{1+q} \frac{q^{-2s}}{1-q^{-2s}}.$$ Here the integral converges for $s>0$ and this equation gives an analytic continutation for $s<0$, and this expression is negative for $s<0$. For the standard spherical vector $\phi_K$, we have $$\langle \phi_K, \phi_K\rangle = \frac{1-q^{-1-2s}}{1-q^{-2s}}$$ which is negative if $-\frac{1}{2} < s < 0$ but positive $s<-\frac{1}{2}$. So it can’t be unitarizable for $s< -\frac{1}{2}$ (since it is not positive definite) and it remain to show that it is unitarizable for $-\frac{1}{2} < s< 0$. To prove this, we consider a new intertwining operator $M_{s}^{*} = (1-q^{-2s})M_{s}$. The original $M_{s}$ is not defined at $s = 0$ (it has a pole), but $M_{s}^{*}$ is even defined at $s = 0$ and it varies continuously. We know that $\mathcal{B}(\chi_s, \chi_s^{-1})$ is unitary for $s = 0$, and the new pairing $\langle f_1, f_2\rangle^{*} = (1-q^{-2s})\langle f_1, f_2\rangle$ become positive definite. Now let $\rho$ be an irreducible admissible representation of $K$. If it is not positive definite for some $-\frac{1}{2} <s < 0$, then there exists $s$ such that $M_{s}^{*}$ has zero eigenvalue, which means that $M_{s}^{*}$ is not invertible. This contradicts to the fact that $M_{s}$ is nonzero, so invertible for $-\frac{1}{2} <s < 0$. Hence $\langle \,,\,\rangle^{*}$ defines a positive definite Hermitian form on $V(\rho)$, so on $V = \oplus_{\rho \in \widehat{K}} V(\rho)$. ◻

Proof. Since it is trivial if $J(V) = 0$, assume that $J(V) \neq 0$. By Theorem \ref{jacadm}, $J(V)$ is admissible $T(F)$-module, so is its contragredient. Since $T(F)$ is abelian, there exists 1-dimensional $T(F)$-invariant subspace of $J(V)^{*}$, which means that there exists a quasicharacter $\chi$ of $T(F)$ and a nonzero linear functional $L:J(V)\to \mathbb{C}$ such that $L(\pi_{N}(t)v) = (\delta^{1/2}\chi)(t)L(v)$ for $v\in J(V), t\in T(F)$. If we consider $L$ as a linear functional on $V$ that is trivial on $V_N$, we have $L(\pi(b)v) = (\delta^{1/2}\chi)(b)L(v)$ for $v\in V, b\in B$. (Here we extend $\chi$ to $B(F)$ that is trivial on $N(F)$. By Frobeinus reciprocity, this corresponds to a nonzero intertwining map $V \to \mathcal{B}(\chi_1, \chi_2)$, which is injective because of irreducibility of $V$. Since Jacquet functor is exact, we have $J(V)\hookrightarrow J(\mathcal{B}(\chi_1, \chi_2))$ and the result follows from $\dim J(\mathcal{B}(\chi_1, \chi_2)) = 2$. ◻

Proof. Recall that the action of $B_{1}(F)$ on the Kirillov model is given as $$\pi\begin{pmatrix} a & \\ & 1 \end{pmatrix} \phi(x) = \phi(ax), \qquad \pi\begin{pmatrix} 1 & b \\ & 1 \end{pmatrix}\phi(x) = \psi(bx)\phi(x).$$ Since $\pi$ is a smooth representation, $\phi$ is fixed by an open subgroup of $T_1(F)$, and the first equation implies that $\phi$ is locally constant. Also, $\phi$ is fixed by $N(\mathfrak{p}^k)$ for some $k$, which gives $\phi(y) = \psi(xy) \phi(y)$ for all $x\in \mathfrak{p}^k$. If $|y|$ is sufficiently large, then $\psi(xy) \neq 1$ and so $\phi(y) = 0$. For the last claim, a function $\phi' = \pi\left(\begin{smallmatrix} 1 & x \\ & 1 \end{smallmatrix}\right)\phi - \phi$ satisfies $\phi'(y) = (\psi(xy) -1 )\phi(y)$, and if $|y|$ is sufficiently small then $\psi(xy) = 1$, hence $\phi' = 0$. ◻

Proof. By the previous proposition, we have $V_N\subseteq C_{c}^{\infty}(F^{\times})$. Also, $V_N\neq 0$ since $\dim V = \infty$ and $\dim J(V) <\infty$. So it is enough to show that $C_{c}^{\infty}(F^{\times})$ is an irreducible $B_1(F)$-module.

Let $U$ be a nonzero invariant subspace of $C_c^{\infty}(F^{\times})$. For any $a\in F^{\times}$, we will show that $U$ contains a characteristic function of any sufficiently small neighborhood of $a$, which proves $U = C_{c}^{\infty}(F^{\times})$. Let $0\neq \phi\in U$ and we may assume $\phi(a)\neq 0$. Let $W$ be an open neighborhood of $a$ such that $\phi|_{W} = \phi(a)$. Choose $f\in C_{c}^{\infty}(F)$ such that $\widehat{f} = \frac{1}{f(a)} \mathds{1}_{W}$, then $$\phi_1(y) := \int\displaylimits_{F} f(x) \pi\begin{pmatrix} 1 & x \\ & 1 \end{pmatrix}\phi(y) dx = \int\displaylimits_{F}f(x)\psi(xy)\phi(y)dx = \widehat{f}(y)\phi(y) = \mathds{1}_{W}(y)$$ is in $C_{c}^{\infty}(F^{\times})$ since it is a finite sum of elements of $C_{c}^{\infty}(F^{\times})$. ◻

Proof. Let $t_0 \in \varpi\mathcal{O}_F^{\times}$. By assumption, $$\pi\begin{pmatrix} t_0 & \\ & 1 \end{pmatrix}\phi - (\delta^{1/2}\chi)\begin{pmatrix} t_0 & \\ & 1 \end{pmatrix} \phi$$ is in $V_N$, so it vanishes near zero. So there exists a constant $\epsilon(t_0)>0$ such that $$\phi(tu) - |t|^{1/2}\chi_1(t)\phi(u) = 0$$ for $t = t_0$ and $|u|\leq \epsilon(t_0)$. By smoothness of $\pi$ and $\chi$, it also holds when $t$ is near $t_0$. By compactness of $\varpi \mathcal{O}_F^{\times}$, there exists uniform $\epsilon >0$ such that the above equation is true for $t\in \varpi \mathcal{O}_{F}^{\times}$ and $|u|\leq \epsilon$. By factoring $0\neq t\in \mathfrak{p}$ as a product of elements of $\varpi \mathcal{O}_{F}^{\times}$, we get the same equation for $0\neq t\in \mathfrak{p}$ and $|u|\leq \epsilon$, which proves the claim. ◻

Proof. This follows from Theorem \ref{psjac} and the previous proposition. For details, see p.515 of (bu?). ◻

Proof. The proof is almost same as the proof of Theorem \ref{pskr}. Note that $T(F)$ acts as $\delta^{1/2}\chi$ on the Jacquet module of $\sigma(\chi_1, \chi_2)$. ◻

Proof. We will only show the case where $(\pi, V)$ is a spherical principal series representation and $\xi$ is unramified. By Theorem \ref{pskr}, we can assume that $\phi(y) = 0$ for $|y| > q^N$ and $\phi(y) = C_{1}|y|^{1/2}\chi_1(y) + C_{2}|y|^{1/2}\chi_2(y)$ for $|y|\leq q^{-N'}$. Then the integral can be written as \begin{align*}Z(s, \phi, \xi) &= \sum_{m\in \mathbb{Z}} \int_{|y| = q^{m}} \phi(y)\xi(y) |y|^{s-1/2} d^{\times}y \\ &= \sum_{m=-N}^{N' - 1} q^{-m(s-1/2)} \int\displaylimits_{|y| = q^{-m}} \phi(y)\xi(y)d^{\times}y \\ &+ \sum_{m\geq N'} q^{-m(s-1/2)}\int\displaylimits_{|y| = q^{-m}} (C_{1}|y|^{1/2}\chi_1(y) + C_{2}|y|^{1/2} \chi_2(y)) \xi(y) d^{\times}y \\ &=r_{\phi}(q^{-s}) + \sum_{m\geq N'} [C_{1} (\alpha_1 q^{-s})^{m} + C_{2}(\alpha_{2} q^{-s})^{m}]\int\displaylimits_{|y| = q^{-m}}\xi(y)d^{\times}y \\ &= r_{\phi}(q^{-s}) + \sum_{m\geq N'} [C_{1} (\alpha_1 \xi(\varpi) q^{-s})^{m} + C_{2}(\alpha_{2} \xi(\varpi)q^{-s})^{m}] \\ &= r_{\phi}(q^{-s}) + \frac{C_1 \alpha_{1}^{N'} q^{-N's}}{1-\alpha_{1}\xi(\varpi)q^{-s}} + \frac{C_2 \alpha_{2}^{N'} q^{-N's}}{1-\alpha_{2}\xi(\varpi)q^{-s}} \\ &=p_{\phi}(q^{-s}) L(s, \pi, \xi)\end{align*} where $p_{\phi}$ is a rational function, $C' = \int_{|y| = 1} \xi(y)d^{\times} y$ and $L(s, \pi, \xi) = L(s, \pi\otimes \xi)$ with $\pi\otimes \xi \simeq \pi(\xi \chi_1, \xi \chi_2)$. Now, define $\phi :F^{\times} \to \mathbb{C}$ as $$\phi(y) = \begin{cases} \frac{\alpha_{1}}{\alpha_{1} - \alpha_{2}} |y|^{1/2} (\xi\chi_{1})(y) - \frac{\alpha_{2}}{\alpha_{1} - \alpha_{2}} |y|^{1/2} (\xi\chi_{2})(y) & y\in \mathcal{O}_F \\ 0 & \text{otherwise}\end{cases}$$ Then this function is in the Kirillov model of $\pi(\xi \chi_1, \xi\chi_2)$, and we can check $Z(s, \phi, \xi) = L(s, \pi, \xi)$ by the above computation. ◻

Proof. For fixed $s$, define two linear functionals $L_1, L_2$ on $V$ by $$L_1(\phi) = Z(s, \phi, \xi), \qquad L_2(\phi) = Z(1-s, \pi(w_1)\phi, \omega^{-1}\xi^{-1}).$$ We can check that both linear functionals satisfy $$L\left( \pi\begin{pmatrix} y & \\ & 1 \end{pmatrix} \phi\right) = \xi(y)^{-1}|y|^{-s+1/2} L(\phi)$$ by using change of variables and analytic continuations. Using Proposition \ref{distuniq}, one can show that $L_{1}$ and $L_{2}$ are linearly dependent when restricted to $V_N$, so $c_1 L_1 + c_2 L_2$ factors through $J(V)$ for some $c_1, c_2\in \mathbb{C}$ that not both zero. This implies that $c_1 L_1 + c_2 L_2 = 0$ for all but two possible choices of $s$ in $\mathbb{C}$ modulo $2\pi i / \log (q)$, and meromorphic continuation proves that they are proportional for all $s$, i.e. there exists a meromorphic function $\gamma(s, \pi, \xi, \psi)$ satisfies $L_{1} = \gamma L_{2}$. ◻

Proof. Let’s identify $\pi_1, \pi_2$ with their Kirillov models, so $V_1, V_2$ are subspaces of $C^{\infty}(F^{\times})$ with $\mathrm{GL}(2, F)$-actions. Let $V_0 = V_1 \cap V_2$. It is enough to show that $\pi_1(w_1)\phi = \pi_2(w_1)\phi$ for $\phi\in V_0$, because this implies $V_1, V_2$ are nonzero irreducible $\mathrm{GL}(2, F)$-spaces so we get $V_1= V_0 = V_2$. (Note that $B(F)$ and $w_1$ generate $\mathrm{GL}(2, F)$.) Also, if we put $\phi_i = \pi_i(w_1)\phi$, then it is sufficient to show that $\phi_1(1) = \phi_2(1)$ by considering the action of $\left(\begin{smallmatrix} a & \\ & 1 \end{smallmatrix}\right)$ for $a\in F^{\times}$. To show this, we use Fourier inversion formula: if $M$ is compact abelian group with normalized Haar measure (so that $|M| = 1$), and if $F$ is a continuous function on $M$, then $$F(1) = \sum_{\chi\in \hat{M}} \,\,\int\displaylimits_{M}F(m)\chi(m)dm.$$ Now for $n\in \mathbb{Z}$, let $$F_{\xi}(n) = \int\displaylimits_{|y| = q^{-n}} (\phi_1(y) - \phi_2(y))\xi(y) d^{\times}y.$$ Then $F_{\xi}(0)$ depends only on the restriction of $\xi$ on $\mathcal{O}_{F}^{\times}$, and $F_{\xi}(0) = 0$ for all but finitely many characters $\xi$ of $\mathcal{O}_{F}^{\times}$ and $$\phi_{1}(1) - \phi_{2}(1) = \sum_{\xi\in \widehat{\mathcal{O}_{F}^{\times}}} F_{\xi}(0)$$ by Fourier inversion formula applied to $M =\mathcal{O}_{F}^{\times}$ and $F(y) = \phi_{1}(y) - \phi_{2}(y)$. By hypothesis and the functional equations of local zeta functions, we have $Z(s, \phi_1, \xi) = Z(s, \phi_2, \xi)$ for all characters $\xi$ of $F^{\times}$. Also, since $\phi_i(y) = 0$ for sufficiently large $|y|$, $F_{\xi}(n) =0$ for sufficiently small $n$. If we put $x = q^{-s +1/2}$, then $$\sum_{n\in \mathbb{Z}} F_{\xi}(n) x^{n} = Z(s, \phi_1, \xi) - Z(s, \phi_2, \xi) = 0$$ for sufficiently small $x$ (i.e. sufficiently large $\Re s$), so $F_{\xi}(n) = 0$ for all $n$, and in particular, $F_{\xi}(0) = 0$. ◻

Proof. The proof uses Whittaker model of $\mathcal{B}(\chi_1, \chi_2)$ constructed in section 3.6 using the Weil representation. Indeed, we defined another (and simpler) Whittaker model in the proof of Proposition \ref{intcomp}. However, the one constructed by means of Weil representation is much more helpful to prove this. It allows us to express local zeta integral of the principal series representation as a product of two local zeta integrals corresponds to two quasicharacters $\xi\chi_1$ and $\xi\chi_2$ naturally, and the result directly follows from this. For details, see p. 548 of (bu?). ◻

Proof. By no small subgroup argument (Proposition \ref{nss}), $\ker (\chi|_{\mathbb{A}_{\mathrm{fin}}^{\times}})$ contains an open neighborhood of the identity. ◻

Proof. Let $N$ be a positive integer, and let \begin{align*}S_{0}(N) &= \{v\,:\, v\text{ is non-archimedean},\, p_{v}|N\} \\ S_{1}(N) &= \{v\,:\, v\text{ is non-archimedean},\, p_{v}\nmid N\}.\end{align*} For each $v\in S_{0}(N)$, let $$U_{v}(N) = \{x\in\mathcal{O}_{v}\,:\, x\equiv 1\,(\mathrm{mod}\,N)\}$$ and $$U_{\mathrm{fin}}(N) = \prod_{v\in S_{0}(N)} U_{v}(N) \times \prod_{v\in S_{1}(N)} \mathcal{O}_{v}^{\times}, \quad U(N) = \mathbb{R}_{+}^{\times} \times U_{\mathrm{fin}}(N).$$ Then $U_{\mathrm{fin}}(N)$ form a basis of neighborhoods of the identity in $\mathbb{A}_{\mathrm{fin}}^{\times}$, so there exists $N$ such that $\chi|_{U_{\mathrm{fin}}(N)}=1$ by no small subgroup argument. The restriction $\chi|_{\mathbb{R}_{+}^{\times}}$ is of the form $|x|^{\lambda}$ for some unique $\lambda\in i\mathbb{R}$, so $\chi_{1}(x):= \chi(x)|x|^{-\lambda}$ is trivial on $U(N)$. If we put $$V(N) = \mathbb{R}_{+}^{\times} \times \prod_{v\in S_{0}(N)} U_{v}(N) \times\sideset{}{'}\prod_{v\in S_{1}(N)} \mathbb{Q}_{v}^{\times},$$ then this is an open subgroup and $\mathbb{A}^{\times} = \mathbb{Q}^{\times}V(N)$ by the approximation theorem. Hence $\mathbb{A}^{\times}/\mathbb{Q}^{\times} \simeq V(N) /(\mathbb{Q}^{\times}\cap V(N))$ and it is enough to show that $\chi_{1}|_{V(N)}$ has finite order. Since $\chi_{1}$ is trivial on $U(N)(\mathbb{Q}^{\times}\cap V(N))$, it is enough to show $[V(N):U(N)(\mathbb{Q}^{\times} \cap V(N))] <\infty$. In fact, we have $$V(N)/U(N)(\mathbb{Q}^{\times}\cap V(N)) \simeq I_{N}/P_{N} \simeq (\mathbb{Z}/N\mathbb{Z})^{\times}$$ where $I_{N}$ is a group of all fractional ideals of $\mathbb{Q}$ prime to $N$, and $P_{N}$ is the subgroup of principal fractional ideals $\alpha\mathbb{Z}$ with $\alpha\in \mathbb{Q}^{\times} \cap V(N)$.

2 follows from composing with the above isomorphism. Note that we have to take minimal $N$ to make the corresponding Dirichlet character primitive. ◻

Proof. For $t\in \mathbb{A}^{\times}$, define $$F(x) = \sum_{\alpha\in F} \Phi((x+\alpha) t).$$ This is a continuous function on the compact abelian group $\mathbb{A}/F$ and has a Fourier expansion $$F(x) = \sum_{\beta\in F} c_{\beta} \psi(-\beta x).$$ By orthogonality of characters, coefficients can be computed by \begin{align*}c_{\beta} &= \frac{1}{V} \int_{\mathbb{A}/F} F(x) \psi(\beta x)dx \\ &= \frac{1}{V} \int_{\mathbb{A}/F} \sum_{\alpha\in F} \Phi((x+\alpha)t) \psi(\beta(x+\alpha)) dx \\ &= \frac{1}{V} \int_{\mathbb{A}} \Phi(xt)\psi(\beta x)dx \\ &=\frac{1}{V|t|} \int_{\mathbb{A}} \Phi(x)\psi(\beta x/t) dx = \frac{1}{V|t|} \widehat{\Phi}\left(\frac{\beta}{t}\right).\end{align*} Here $V$ is the volume of $\mathbb{A}/F$ and we use the substitution $x \to x/t$ in the last equality. Now put $x =0$ and we get $$\sum_{\alpha\in F} \Phi(\alpha t) = F(0) = \sum_{\beta\in F} c_{\beta}= \frac{1}{V|t|} \sum_{\beta\in F}\widehat{\Phi}\left(\frac{\beta}{t}\right).$$ If we apply this twice and put $t = 1$, then $$\sum_{\alpha\in F} \Phi(\alpha) = \frac{1}{V^{2}} \sum_{\alpha\in F}\Phi(-\alpha)$$ and this implies $V = 1$. ◻

Proof. First, we will show that the integral absolutely converges for $\Re s > 0$. Since $\chi_{v}$ is unitary, the integral is bounded by $$\int_{F_{v}^{\times}} |\Phi_{v}(x)| |x|_{v}^{s} d^{\times}x_{v} = \int_{|x|_{v} \leq 1}|\Phi_{v}(x)| |x|_{v}^{s} d^{\times}x_{v} + \int_{|x|_{v} > 1} |\Phi_{v}(x)| |x|_{v}^{s} d^{\times}x_{v}$$ For the above two integrals on RHS, second integral absolutely converges because of rapid decay of $\Phi_{v}(x)$ (the function is in the Schwartz space). For the first integral corresponds to the region $|x|_{v} \leq 1$, $|\Phi_{v}(x)|$ is bounded because $|x|_{v} \leq 1$ is compact. So we can ignore $\Phi_{v}(x)$ and the remaining term decomposes as $$\sum_{k\geq 0} \int_{\mathrm{ord}_{v}(x) = k} |x|_{v}^{s} d^{\times}x_{v} = \sum_{k\geq 0} q_{v}^{-ks}$$ when $v$ is non-archimedean, and the summation converges for $\Re s >0$. Real and complex case comes from the convergence of the integrals $$\int_{-1}^{1} |t|^{s} \frac{dt}{t}, \quad \frac{1}{2\pi} \int_{0}^{2\pi} \int_{0}^{1} r^{2s} \frac{dr}{r} d\theta$$ for $\Re s >0$.

Now let $S$ be a finite set of primes so that for all $v\not\in S$, $v$ is non-archimedean, $\chi_{v}$ is unramified and $\Phi_{v} = \mathds{1}_{\mathcal{O}_{v}}$. Then \begin{align*}\zeta_{v}(s, \chi_{v}, \Phi_{v}) &= \int_{F_{v}^{\times}} \mathds{1}_{\mathcal{O}_{v}}(x) \chi_{v}(x) |x|_{v}^{s} d^{\times}x_{v} \\ &= \sum_{k\geq 0} \int_{\mathrm{ord}_{v}(x) = k} \chi_{v}(\mathfrak{p}_{v})^{k} q_{v}^{-ks} d^{\times}x_{v} \\ &= \sum_{k\geq 0} (\chi_{v}(\mathfrak{p}_{v})q_{v}^{-s})^{k} = (1-\chi_{v}(\mathfrak{p}_{v})q_{v}^{-s})^{-1}\end{align*} Now the product $$\prod_{v} \zeta_{v}(s, \chi_{v}, \Phi_{v})$$ absolutely converges for $\Re s > 1$ because the local zeta integrals agrees with the local factors of corresponding Hecke $L$-function. ◻

Proof. First, we show that $\gamma$ doesn’t depend on the test function when $0<\Re s < 1$. In other words, we must prove that $$\zeta_{v}(1-s, \chi_{v}^{-1}, \widehat{\Phi}_{v}) \zeta_{v}(s, \chi_{v}, \Phi_{v}') = \zeta_{v}(1-s, \chi_{v}^{-1}, \widehat{\Phi_{v}'}) \zeta_{v}(s, \chi_{v}, \Phi_{v}).$$ The LHS equals to \begin{align*}&\int_{F_{v}^{\times}}\left[ \int_{F_{v}} \Phi_{v}(y) \psi_v(xy)dy\right]\chi_{v}(x)^{-1}|x|_{v}^{1-s} d^{\times}x \int_{F_{v}^{\times}} \Phi_{v}'(z)\chi_{v}(z) |z|_{v}^{s} d^{\times}z \\ &= \int_{F_{v}^{\times}}\int_{F_{v}}\int_{F_{v}^{\times}} \Phi_{v}(y)\Phi'_{v}(z)\psi_v(xy)\chi_{v}(x)^{-1}\chi_{v}(z) |x|_{v}^{1-s}|z|_{v}^{s} d^{\times}x dyd^{\times}z \\ &= \frac{1}{m_{v}} \int_{F_{v}^{\times}}\int_{F_{v}^{\times}}\int_{F_{v}^{\times}} \Phi_{v}(y)\Phi'_{v}(z) \chi_{v}(yz) |yz|_{v}^{s} \psi_v(x) \chi_{v}(x)^{-1} |x|_{v}^{1-s}d^{\times}x d^{\times}y d^{\times} z\end{align*} where the last equality follows from the substitution $x\to y^{-1}x$ for $y\neq 0$. Here $m_{v}$ is a constant that satisfies $d^{\times}y = m_{v}dy / |y|_{v}$ between nomarlized additive Haar measure and normalized multiplicative Haar measure. Clearly, this is symmetric in $\Phi_{v}$ and $\Phi_{v}'$, so the equation holds.

To extend this for all $s$, choose $\Phi_{v}$ so that $\widehat{\Phi}_{v}$ vanishes in a neighborhood of zero. Then $\zeta_{v}(1-s,\chi_{v}^{-1}, \widehat{\Phi}_{v})$ is convergent for all $s$ and we already know that $\zeta_{v}(s, \chi_{v}, \Phi_{v})$ is convergent and holomorphic for $\Re s >0$. This implies that their quotient $\gamma_{v}(s, \chi_{v}, \psi_{v})$ has a meromorphic continuation on $\Re s> 0$, and we get a similar result for $\Re s <1$ by choosing nice $\Phi_{v}$ that vanishes near zero. This also proves 1 together with the previous proposition.

For 3, choose $\Phi_{v}$ as compactly supported near $x = 1$, so that integral converges for any $s\in \mathbb{C}$ and $\chi_{v}(x) |x|^{s_{0}}_{v}$ has positive real part on the support of $\Phi_{v}$, so is nonzero. For non-archimedean $v$, choose $\Phi_{v}$ so that the support of $\Phi_{v}$ is in $\mathcal{O}_{v}^{\times}$, then $\zeta_{v}(s, \chi_{v}, \Phi_{v})$ is independent of $s$ and we can normalize it as 1. ◻

Proof. The main idea is to split the integral into two parts, $|x| <1$ and $|x|>1$, and using the Poisson summation formula to unfold and refold integral. Let \begin{align*}\zeta_{1}(s, \chi, \Phi) &:= \int\displaylimits_{\substack{\mathbb{A}^{\times} \\ |x| > 1}} \Phi(x)\chi(x)|x|^{s} d^{\times} x \\ \zeta_{0}(s, \chi, \Phi) & := \int\displaylimits_{\substack{\mathbb{A}^{\times} \\ |x| < 1 }} \Phi(x) \chi(x) |x|^{s} d^{\times}x\end{align*} We already know that the first integral converges for all $s$, since it converges for $\Re s > 1$ by the previous results and decreasing $\Re s$ only improves the convergence in the region $|x| >1$. For the second one, we can unfold the integral as \begin{align*}\zeta_{0}(s, \chi, \Phi) &= \sum_{\alpha\in F^{\times}} \int\displaylimits_{\substack{\mathbb{A}^{\times}/F^{\times} \\ |x| < 1}} \Phi(\alpha x)\chi(\alpha x) |\alpha x|^{s} d^{\times}x \\ &= \int\displaylimits_{\substack{\mathbb{A}^{\times}/F^{\times} \\ |x| < 1}} \left[\sum_{\alpha\in F^{\times}} \Phi(\alpha x)\right] \chi(x) |x|^{s} d^{\times}x \\ &= \int\displaylimits_{\substack{\mathbb{A}^{\times}/F^{\times} \\ |x| < 1}} \left[\sum_{\alpha\in F} \Phi(\alpha x)\right] \chi(x) |x|^{s} d^{\times}x - \Phi(0) \int\displaylimits_{\substack{\mathbb{A}^{\times}/F^{\times} \\ |x|<1}} \chi(x)|x|^{s}d^{\times}x\end{align*} since $\chi(\alpha) = 1 = |\alpha|$ for all $\alpha\in F$. The last integral can be written as $$\int\limits_{0}^{1}\int\displaylimits_{\substack {\mathbb{A}^{\times}/F^{\times} \\ |x| = t}} \chi(x)|x|^{s}d^{\times}x \frac{dt}{t}$$ If $\chi|_{\mathbb{A}^{\times}_{1}}$ is nontrival, then this integral is zero since there exists $y_{0}\in \mathbb{A}^{\times}_{1}$ with $\chi(y_{0}) \neq 1$ and $$\int\limits_{\substack{\mathbb{A}^{\times}/F^{\times} \\ |x| = t}} \chi(x)|x|^{s} d^{\times}x = \int\limits_{\substack{\mathbb{A}^{\times}/F^{\times} \\ |x| = t}} \chi(xy_{0})|xy_{0}|^{s} d^{\times}x = \chi(y_{0})\int\limits_{\substack{\mathbb{A}^{\times}/F^{\times} \\ |x| = t}} \chi(x)|x|^{s} d^{\times}x$$ impiles that the integral vanishes. If it is trivial, then $\chi(x) = |x|^{\lambda}$ for some $\lambda\in i\mathbb{R}$ (since $\chi$ is unitary) and the integral became $$\rho_{F} \int_{0}^{1} t^{s+\lambda} \frac{dt}{t} = \frac{\rho_{F}}{s+\lambda}$$ where $\rho_{F}$ is the volume of $\mathbb{A}^{\times}_{1} /F^{\times}$. By the Poisson summation formula, we have \begin{align*}\zeta_{0}(s, \chi, \Phi) = \begin{cases} \int\displaylimits_{\substack{\mathbb{A}^{\times}/F^{\times} \\ |x| < 1}} \left[ \sum_{\alpha\in F} \widehat{\Phi}\left(\frac{\alpha}{x}\right)\right] \chi(x) |x|^{s-1}d^{\times}x & \chi|_{\mathbb{A}^{\times}_{1}} \neq 1 \\ \int\displaylimits_{\substack{\mathbb{A}^{\times}/F^{\times} \\ |x| < 1}} \left[ \sum_{\alpha\in F} \widehat{\Phi}\left(\frac{\alpha}{x}\right)\right] \chi(x) |x|^{s-1}d^{\times}x - \frac{\rho_{F}\Phi(0)}{s + \lambda}& \chi|_{\mathbb{A}^{\times}_{1}} = 1\end{cases}\end{align*} Now apply the change of variable $x\to x^{-1}$ and we get \begin{align*}\zeta(s, \chi, \Phi) = \begin{cases}\zeta_{1}(s, \chi, \Phi) + \zeta_{1}(1-s, \chi^{-1}, \widehat{\Phi}) & \chi|_{\mathbb{A}^{\times}_{1}} \neq 1 \\ \zeta_{1}(s, \chi, \Phi) + \zeta_{1}(1-s, \chi^{-1}, \widehat{\Phi}) - \rho_{F} \left( \frac{\Phi(0)}{s+\lambda} + \frac{\widehat{\Phi}(0)}{1-s-\lambda}\right)& \chi|_{\mathbb{A}^{\times}_{1}} = 1\end{cases}\end{align*} Essential boundedness follows from $$|\zeta_{1}(s, \chi, \Phi)| \leq \int\displaylimits_{\substack{\mathbb{A}^{\times} \\ |x| >1 }} |\Phi(x)||x|^{\Re (s)} d^{\times} x.$$ ◻

Proof. We proved that $L_{v}(s, \chi_{v}) = \zeta_{v}(s, \chi_{v}, \Phi_{v})$ holds except for finitely many places (especially, for $v\not\in S$). Then the functional equation of the local and global zeta integrals give the result. ◻

Proof. For a non-archimedean $v$, we can write the local zeta integral as $$\zeta_{v}(s, \chi_{v}, \Phi_{v}) = \sum_{k\in \mathbb{Z}} q_{v}^{ks} \int\limits_{|x|_{v} = q_{v}^{k}} \Phi_{v}(x)\chi_{v}(x)d^{\times}x.$$ Since $\Phi_{v}(x)$ is compactly supported, the contribution is zero for large $k$. Also, if $-k$ is large, then $\Phi_{v}(x) = \Phi_{v}(0)$ (since the function is locally constant) and the contribution equals $$\Phi_{v}(0)\int\limits_{|x|_{v} = q_{v}^{k}} \chi_{v}(x)d^{\times}x.$$ If $\chi_{v}$ is ramified, then this is zero and the sum is only a finite sum, i.e. $\zeta_{v}(s, \chi_{v}, \Phi_{v})$ is entire and rational in $q_{v}^{-s}$. If $\chi_{v}$ is unramified, then it became a geometric series for small $k$, and we can explicitly describe pole of the function.

If $v$ is real, the integral can be decomposed as integral over $|x|\leq 1$ and $|x|>1$ as before, and we only need to analyze the first part since the second part converges absolutely for all $s$ by rapid decay of $\Phi_{v}$. We may split $\Phi_{v}$ into even and odd parts and handle these two cases separately. The integral vanishes unless the parity of $\Phi_{v}$ and $\chi_{v}$ matches. If $\chi_{v} = 1$ and $\Phi_{v}$ is even, then the Taylor expansion of $\Phi_{v}(x)$ has only even terms and the possible poles of the integral $$\int\limits_{|x|\leq 1} \Phi_{v}(x) \chi_{v}(x)|x|_{v}^{s} d^{\times}x = 2\int_{0}^{1} \Phi_{v}(x) x^{s} d^{\times} x$$ are at $s = 0, -2, -4, \dots$, which agrees with the poles of $L_{v}(s, \chi_{v})$. Similarly thing holds for odd cases, too.

If $v$ is complex, we use polar coordinate and $$\zeta_{v}(s,\chi_{v}, \Phi_{v}) = \int_{0}^{\infty} r^{2\nu + 2s} \phi(r) \frac{dr}{r}$$ where $$\phi(r) = \frac{1}{2\pi} \int_{0}^{2\pi} \Phi_{v}(re^{ik\theta}) e^{ik\theta}d\theta.$$ We consider the Taylor expansion of $\Phi_{v}$ and $\phi(r)$, which gives $$\phi(r) = \sum_{m-n = k} a(n, m) r^{n+m}$$ where $a(n, m)$ is a Taylor coefficient of $\Phi_{v}(x)$ of $x^{n}\overline{x}^{m}$. Then we get the result about poles by the same argument as real case.

For the suitable choice of $\Phi_{v}$, choose \begin{align*}\Phi_{v}(x) = \begin{cases} \mathds{1}_{\mathcal{O}_{v}}(x) & v\text{ non-archimedean, $\chi_{v}$ unramified} \\ \chi_{v}(x)^{-1} \mathds{1}_{\mathcal{O}_{v}^{\times}}(x) & v\text{ non-archimedean, $\chi_{v}$ ramified} \\ x^{\epsilon}e^{-\pi x^{2}} & v \text{ real} \\ \overline{x}^{k} e^{-2\pi |x|_{v}} & v\text{ complex}, k>0 \\ x^{-k} e^{-2\pi |x|_{v}} & v \text{ complex}, k<0 \end{cases}\end{align*} then we get $L_{v}(s, \chi_{v}) = \zeta_{v}(s, \Phi_{v}, \chi_{v})$ for all $v$. ◻

Proof. By definition, we have $$\epsilon_{v}(s, \chi_{v}, \psi_{v}) = \frac{\zeta_{v}(1-s, \chi_{v}^{-1}, \widehat{\Phi}_{v})}{L_{v}(1-s, \chi_{v}^{-1})} \cdot \frac{L_{v}(s, \chi_{v})}{\zeta_{v}(s, \chi_{v}, \Phi_{v})}$$ and the result follows from the previous proposition. ◻

Proof. The functional equation follows from previous propositions: \begin{align*}L(s, \chi) &=\left( \prod_{v\in S}L_{v}(s, \chi_{v})\right) L_{S}(s, \chi) \\ &= \left(\prod_{v\in S} L_{v}(s, \chi_{v}) \gamma_{v}(s, \chi_{v}, \psi_{v}) L_{v}(s, \chi_{v})\right) L_{S}(1-s, \chi^{-1}) \\ &= \left( \prod_{v\in S} \frac{L_{v}(s, \chi_{v}) \gamma_{v}(s, \chi_{v}, \psi_{v})}{L_{v}(1-s, \chi_{v}^{-1})}\right) L(1-s, \chi^{-1}) \\ &= \epsilon(s, \chi) L(1-s, \chi^{-1})\end{align*} and it is evident from the functional equation that $\epsilon(s, \chi)$ is independent of the choice of $\psi$. Essential boundedness follows from $$L(s, \chi) = \zeta(s, \Phi, \chi) \prod_{v} \frac{L_{v}(s, \chi)}{\zeta_{v}(s, \Phi_{v}, \chi_{v})}$$ and the fact that we can choose $\Phi = \prod_{v}\Phi_{v}$ so that the quotient $L_{v} / \zeta_{v}$ is exponential type for all $v$ (and identically 1 for almost all $v$), and $\zeta(s, \Phi, \chi)$ is essentially bounded. ◻

Proof. This is a theorem of Harish-Chandra. More precisely, he proved the same theorem for $G = \mathrm{SL}(2, \mathbb{R})$ and $K = \mathrm{SO}(2)$. Now we can reduce the original statement to the $\mathrm{SL}(2, \mathbb{R})$ case. Indeed, for $f\in \mathcal{A}(\Gamma \backslash G, \chi, \omega)$, $|f|$ is constant on the cosets of $Z(\mathbb{R})$ and it is easy to see that in each coset of $Z(\mathbb{R})$, the element $g$ with the minimal height $||g||$ is actually in $\mathrm{SL}(2, \mathbb{R})$. For the proof of Harish-Chandra’s theorem, see Theorem 2.9.2 of (bu?). ◻

Proof. One can find $c, d>0$ such that $\xi_{i}^{-1}\mathcal{F}_{c, d}$ contains a neighborhood of the cusp $a_i$ in the fundamental domain $F$ of $\Gamma$, and so $F - \bigcup_{i} \xi_{i}^{-1}\mathcal{F}_{c, d}$ is relatively compact in $\mathcal{H}$. Then we can increase $d$ and decrease $c$ so that $F = \bigcup_{i} \xi_{i}^{-1} \mathcal{F}_{c, d}$. For 2, we can also find sufficiently large $d'$ so that $\mathcal{F}_{d'}^{\infty}$ contains each $\xi_{i}^{-1}\mathcal{F}_{c, d}$. ◻

Proof. Recall that $\Gamma$ has cusps. We can assume that $\infty$ is a cusp of $\Gamma$ and that $\Gamma$ contains the group $$\Gamma_{\infty} = \left\{ \begin{pmatrix} 1 & n \\ & 1 \end{pmatrix} \,:\, n\in \mathbb{Z}\right\}.$$ By the Proposition \ref{siegel}, it is enough to show that $$\sup_{g\in \mathcal{G}_{c, d}}|\rho(\phi)f(g)| \leq C_{0}||f||_{2}$$ for some $C_{0} >0$. We have \begin{align*}(\rho(\phi)f)(g) &= \int_{Z(\mathbb{R})\backslash G} f(gh)\phi_{\omega}(h) dh \\ &= \int_{Z(\mathbb{R})\backslash G} f(h) \phi_{\omega}(g^{-1}h)dh \\ &= \int\limits_{\Gamma_{\infty} Z(\mathbb{R})\backslash G} \sum_{\gamma\in \Gamma_{\infty}} f(\gamma h)\phi_{\omega}(g^{-1}\gamma h) \\ &= \int\limits_{\Gamma_{\infty} Z(\mathbb{R})\backslash G} K(g, h) f(h) dh\end{align*} where $$K(g, h) = \sum_{\gamma\in \Gamma_{\infty}} \chi(\gamma) \phi_{\omega}(g^{-1}\gamma h).$$ Now we will assume $\chi$ is trivial. The nontrivial case is almost same as the following proof. Since $f$ is cuspidal, we have $$\int\limits_{\Gamma_{\infty} Z(\mathbb{R})\backslash G} K_{0}(g, h) f(h) dh = 0$$ where we define $$K_{0}(g, h) = \int_{-\infty}^{\infty} \phi_{\omega} \left( g^{-1} \begin{pmatrix} 1 & x \\ & 1 \end{pmatrix} h\right) dx.$$ So we may write $$(\rho(\phi)f)(g) = \int\limits_{\Gamma_{\infty} Z(\mathbb{R})\backslash G} K'(g, h) f(h) dh$$ where $K'(g, h) = K(g, h) - K_{0}(g, h)$. Now we will estimate this function to get the desired result. By the Poisson summation formula, we have $$K'(g, h) = \sum_{n\neq 0} \widehat{\Phi}_{g, h}(n)$$ where $$\Phi_{g, h}(t) = \phi_{\omega} \left( g^{-1} \begin{pmatrix} 1 & t \\ & 1 \end{pmatrix} h\right).$$ Using Iwasasa decomposition, we can write $g, h$ as $$g = \begin{pmatrix} \eta & \\ & \eta \end{pmatrix}\begin{pmatrix} y & x \\ & 1 \end{pmatrix} \kappa_{\theta}, \quad h = \begin{pmatrix} \zeta & \\ & \zeta \end{pmatrix}\begin{pmatrix} v & u \\ & 1 \end{pmatrix} \kappa_{\sigma}$$ where $y, u, \eta, \zeta > 0$. By the change of variable, we can show that the absolute value of $|\widehat{\Phi}_{g, h}(n)|$ is independent of $x, u$ and $|\widehat{\Phi}_{g, h} (n)| = |y|\,|\widehat{F}_{\theta, \sigma, y^{-1}v}(yn)|$, where $$F_{\theta, \sigma, v}(t) = \phi_{\omega} \left( \kappa_{\theta}^{-1} \begin{pmatrix} 1 & t \\ & 1 \end{pmatrix} \begin{pmatrix} v & \\ & 1 \end{pmatrix} \kappa_{\sigma}\right).$$ (Note that the central character $\omega$ is unitary.) By Fourier theory, since $F$ is smooth, $\widehat{F}$ decays faster than any polynomial, i.e. for any $N$ we have a constant $C = C_{\theta, \sigma, v}$ (vary continuously in $\theta, \sigma, v$) such that $$|\widehat{F}_{\theta, \sigma, v}(y)| \leq C_{\theta, \sigma, v}|y|^{-N}.$$ Since $\phi_{w}$ is compactly supported modulo $Z(\mathbb{R})$, there exists $B>1$ such that $F_{\theta, \sigma, v}(t) = 0$ unless $B^{-1} \leq v \leq B$. So we have $$|\widehat{\Phi}_{g, h}(n)| \leq C_{1} |y|^{1-N}|n|^{-N}$$ where $C_{1} = \max_{(\theta, \sigma, v)\in [0, 2\pi]\times [0, 2\pi] \times [yB^{-1}, yB]} C_{\theta, \sigma, v} < \infty$. Also, $\Phi_{g, h}(n) = 0$ unless $B^{-1} \leq y^{-1}v \leq B$, so by summing up we get $$|K'(g, h)| \leq C_{2} |y|^{1-N}$$ where $N\geq 2$ and $C_{2}$ is a constant depending on $\phi$ and $N$.

To estimate $(\rho(\phi)f)(g)$ for $g\in \mathcal{G}_{c, d}$, since the kernel $K'$ vanishes unless $B^{-1}\leq y^{-1}v$, we have $$|(\rho(\phi)f)(g)| \leq C_{2}y^{1-N} \int\limits_{\overline{\mathcal{G}}_{B^{-1}c, d}}|f(h)| dh.$$ Now $\overline{\mathcal{G}}_{B^{-1}c, d}$ can be covered by a finite number of copies of a fundamental domain, so it is dominated by $L^{1}$ norm of $f$, so is by $L^{2}$ norm (the fundamental domain has finite volume). This proves 1 and also shows that $\rho(\phi)f(g)$ decays rapidly.

To prove compactness of $\rho(\phi)$, we use Arzéla-Ascoli theorem. Let $Y$ be a compactification of $\Gamma Z(\mathbb{R}) \backslash G$ by adjoining cusps and let $\Sigma$ be the image of the unit ball in $L^{2}_{0}(\Gamma\backslash G, \chi, \omega)$ under $\rho(\phi)$; we extend each $\rho(\phi)f$ to $Y$ by making it vanish at the cusps. This $\Sigma$ is bounded because of the inequality we just proved, and equicontinuity follows from that derivatives are bounded uniformly for all $f$ with $||f||_{2} \leq 1$. This follows from $(X\rho(\phi)f)(g) = \rho(\phi_{X})f(g)$ where $$\phi_{X}(g) = \frac{d}{dt} \phi(\exp (-tX)g)|_{t=0}.$$ Hence $\Sigma$ is compact in $L^{\infty}$ topology and hence also in $L^{2}$ topology. ◻

Proof. The proof that $L_{0}^{2}(\Gamma \backslash G, \chi, \omega)$ decomposes into a Hilbert space direct sum of irreducible invariant subspaces is almost same as the proof of Theorem \ref{l2decomp}. Here we use the previous proposition and the spectral theorem for self-adjoint compact operators. Let $H$ be an irreducible invariant subspace of $L_{0}^{2}(\Gamma \backslash G, \chi, \omega)$. Then $H = \oplus_{k} H_{k}$ where $H_{k} = \{f\in H\,:\, \rho(\kappa_{\theta})f = e^{ik\theta}f\}$. By Theorem \ref{mult1char}, $\dim H_{k} \leq 1$ for all $k$, and this implies that $H_{\mathrm{fin}}$ is a $(\mathfrak{g}, K)$-module.

Finally, to show $H_{\mathrm{fin}} \subseteq \mathcal{A}_{0}(\Gamma\backslash G, \chi, \omega)$, it is enough to show that $H_{k} \subseteq \mathcal{A}_{0}(\Gamma\backslash G, \chi, \omega)$. Let $0\neq f\in H_k$. One can prove that there exists $\phi\in C_{c}^{\infty}(G)$ such that $\phi(\kappa_{\theta}g) = \phi(g\kappa_{\theta}) = e^{-ik\theta}\phi(g)$ and $\rho(\phi)f \neq 0$. It is easy to check that $\rho(\phi)f\in H_k$, so from $\dim H_k \leq 1$, we may assume that $\rho(\phi) f = f$. By the way, convolutioning $f$ with a compactly supported smooth function $\phi\in C_{c}^{\infty}(G)$ makes $\rho(\phi)f = f$ as a smooth and rapidly decreasing function (this follows from the equation $X\rho(\phi)f = \rho(\phi_X)f$ and the estimation of $K'(g, h)$), so $f\in \mathcal{A}_{0}(\Gamma \backslash G, \chi, \omega)$. ◻

Proof. We prove 1. Choose $\sigma \in \widehat{K}$ such that $V(\sigma)\neq 0$, and let $0\neq \xi\in V(\sigma)$. One can choose $\phi\in C_{c}^{\infty}(G)$ such that $\pi(\phi)\xi = \xi$ and $\pi(\phi)$ is self-adjoint and compact. If $T:V\to L_{0}^{2}(\Gamma \backslash G, \chi, \omega)$ is an intertwining map, then $\rho(\phi)T\xi = T\xi$, so $T\xi$ lies in the 1-eigenspace of the compact operator $\rho(\phi)$, which is finite dimensional by the spectral theorem. Since $\pi$ is irreducible, $T$ is determined by the image of any single nonzero vector, and it follows that $\mathrm{Hom}_{\mathrm{GL}(2, \mathbb{R})}(V, L_{0}^{2}(\Gamma \backslash G, \chi, \omega))$ is finite dimensional.

For the second part, we assume that $\Delta$ acts as a constant $\lambda$ and also $\omega$ is fixed, so the action of $Z = \left(\begin{smallmatrix} 1 & 0 \\ 0 & 1 \end{smallmatrix}\right)\in \mathcal{Z}(U\mathfrak{g})$ is also given by a constant $\mu$ determined by $\omega$. By Theorem \ref{unigk}, there are only finite number of isomorphism classes of unitary irreducible admissible representations with given $\omega$ and $\lambda$ (in fact, only one or two). Let $\Sigma$ be the set of these isomorphism classes. By the previous Theorem \ref{cuspdecom}, we know that $L_{0}^{2}(\Gamma \backslash G, \chi, \omega)$ decomposes into a direct sum of irreducible admissible representations and that the $K$-finite vectors in each of these are elements of $\mathcal{A}_{0}(\Gamma \backslash G, \chi, \omega)$, so $\mathcal{A}_{0}(\Gamma, \chi, \omega, \lambda, \rho)$ is the direct sum of the $\rho$-isotypic components of the irreducible subspaces of $L_{0}^{2}(\Gamma \backslash G, \chi, \omega)$ that are isomorphic of an element of $\Sigma$. Now the finite dimensionality follows from the finiteness of $\Sigma$ and the finite multiplicity of each representations in $\Sigma$. ◻

Proof. We will only prove when $n = 2$ and $F = \mathbb{Q}$. We can define Siegel sets $\mathcal{G}_{c, d}\subset \mathrm{GL}(2, \mathbb{A})$ as the set of adéles of the form $(g_{v})$ where $g_{v} \in K_v$ for all non-archimedean $v$ and $$g_{\infty} = \begin{pmatrix} z & \\ & z \end{pmatrix} \begin{pmatrix} y & x \\ & 1 \end{pmatrix} \kappa_{\infty}, \qquad z\in \mathbb{R}^{\times}, y\geq c, 0\leq x\leq d, \kappa_{\infty} \in K_{\infty}.$$ Then if $c\leq \sqrt{3}/2$ and $d\geq 1$ we have $\mathrm{GL}(2, \mathbb{A}) = \mathrm{GL}(2, \mathbb{Q}) \mathcal{G}_{c, d}$. This follows from the strong approximation theorem and the fact that $\mathcal{F}_{c, d}$ contains a fundamental domain for $\mathrm{SL}(2, \mathbb{Z})$.

To prove the inequality, we use the same trick as before. We can express $\rho(\phi)f$ as $$(\rho(\phi)f)(g) = \int\displaylimits_{N(F)Z(\mathbb{A})\backslash\mathrm{GL}(2, \mathbb{A})} K'(g, h)f(h)dh,$$ where \begin{align*}K'(g, h) & = K(g, h) - K_{0}(g, h), \\ K(g, h) & = \sum_{\gamma \in N(F)} \phi_{\omega}(g^{-1}\gamma h), \\ K_{0}(g, h) = \int\displaylimits_{\mathbb{A}/F} &\phi_{\omega} \left( g^{-1} \begin{pmatrix} 1 & x \\ & 1 \end{pmatrix} h\right)dx. \end{align*} Let $g\in \mathcal{G}_{c, d}$. We can write it as $$g = \begin{pmatrix} \eta & \\ & \eta \end{pmatrix} \begin{pmatrix} y & x \\ & 1 \end{pmatrix} \kappa_{g}$$ where $\eta\in \mathbb{R}^{\times}, 0\leq x\leq d, y\geq c$ and $\kappa_{g} \in K$. Also, an arbitrary element $h\in \mathrm{GL}(2, \mathbb{A})$ can be written as $$h = \begin{pmatrix} \zeta & \\ & \zeta \end{pmatrix}\begin{pmatrix} v & u \\ & 1 \end{pmatrix} \kappa_{h}$$ where $\zeta v\in \mathbb{A}^{\times}, u\in \mathbb{A}$, and $\kappa_{h} \in K$. By the Poisson summation formula, $$K'(g, h) = \sum_{\alpha\in F^{\times}} \widehat{\Phi}_{g, h}(\alpha)$$ where $\Phi_{g, h}:\mathbb{A}\to \mathbb{C}$ is the compactly supported continuous function $$\Phi_{g, h}(x) =\phi_{\omega} \left( g^{-1} \begin{pmatrix} 1 & x \\ & 1 \end{pmatrix} h\right).$$ By substitution, we can also check that $$\widehat{\Phi}_{g, h}(\alpha) = \psi(\alpha(x-u))\omega(\zeta^{-1}\eta) |y|\widehat{F}_{\kappa_{g}, \kappa_{h}, y^{-1}v}(\alpha y),$$ where $$F_{\kappa_{g}, \kappa_{h}, y}(t) = \phi_{\omega} \left( \kappa_{g}^{-1} \begin{pmatrix} 1 & t \\ & 1 \end{pmatrix} \begin{pmatrix} y & \\ & 1 \end{pmatrix} \kappa_{h}\right).$$ Since $K\mathrm{supp}\,(\phi)K \cap B(\mathbb{R})$ is compact, there exists a compact subset $\Omega \subset \mathbb{A}^{\times}$ such that if $F_{\kappa_g, \kappa_h, y}(t) \neq 0$ for any $t$, then $y\in \Omega$. We have $$|K'(g, h)| \leq |y| \sum_{\alpha\in F^{\times}} |\widehat{F}_{\kappa_g, \kappa_h, y^{-1}v}(\alpha y)|$$ and $F_{\kappa_g, \kappa_h, y^{-1}v}(y)$ vanishing identically unless $(\kappa_g, \kappa_h, y^{-1}v) \in K\times K\times \Omega$, which is a compact set. Therefore for any given $N>0$, there exists a constant $C_N > 0$ such that $|K'(g, h)| \leq C_N |y|^{-N}$ and $K'(g, h) = 0$ unless $y^{-1}v \in \Omega$. Thus $$|(\rho(\phi)f)(g)| \leq C_{N}|y|^{-N} \int\displaylimits_{\mathbb{A}/F} \int\displaylimits_{y^{-1}v\in \Omega}\int\displaylimits_{K} \Bigg| f\left( \begin{pmatrix} v & u \\ & 1 \end{pmatrix} \kappa_{h}\right)\Bigg| |v|^{-1} d\kappa_{h}d^{\times}v du.$$ (Here $|v|^{-1}$ comes from $d_{L}b = |v_{2}/v_{1}| d_{R}b = |v_{2}/v_{1}|d^{\times}v_1 d^{\times}v_2 du$ for $b = \left(\begin{smallmatrix} v_1 & u \\ & v_2 \end{smallmatrix}\right)$.) Since $\mathbb{A}/F \times y\Omega \times K$ is compact, it can be covered by a finite number of copies of a fundamental domain of $\mathrm{GL}(2, F)Z(\mathbb{A})$. Since it has finite measure, RHS is dominated by $||f||_{1}$, so $||f||_{2}$. Compactness of $\rho(\phi)$ follows from Arzela-Ascoli theorem as before, by showing that the image of unit ball in $L^{2}_{0}(\mathrm{GL}(2, F) \backslash\mathrm{GL}(2, \mathbb{A}), \omega)$ is equicontinuous. By no small subgroup argument, there exists an open subgroup of $\mathrm{GL}(2, \mathbb{A}_{\mathrm{fin}})$ under which $\phi$ is right invariant, and any element of the image of $\rho(\phi)$ will be right invariant under this same subgroup. So we only need to show that $(\rho(\phi)f)(g)$ are equicontinuous as functions of $g_{\infty}$, and this follows from a uniform bound for $X\rho(\phi)f$ as before. ◻

Proof. We only prove for $n = 2$ and $F = \mathbb{Q}$. We will reduce this to the Theorem \ref{autofin}. We need to show that $\dim V(\rho) < \infty$ for any irreducible finite dimensional representation $\rho$ of $K$. One can show that $\rho$ decomposes as a restricted tensor product of local factors, i.e. $\rho = \otimes_{v} \rho_{v}$ where $(\rho_{v}, V_{v})$ are finite dimensional representations of $K_v$ and $\rho_v$ is trivial for almost all $v$. (Look up the next section for the definition of restricted product of representations and the proof of this property.)

From this, there exists an open subgroup $K_{0, \mathrm{fin}}\subseteq K_{\mathrm{fin}}$ such that every vector in $V(\rho)$ is invariant under $K_{0, \mathrm{fin}}$ so that $V(\rho) = V^{K_{0,\mathrm{fin}}}(\rho)$.

Now, we will show that $V^{K_{0, \mathrm{fin}}}(\rho_\infty)$ has a finite dimension. Since $V(\rho) = V^{K_{0, \mathrm{fin}}}(\rho) \subseteq V^{K_{0, \mathrm{fin}}}(\rho_{\infty})$, this shows $\dim V(\rho) <\infty$. In fact, we will prove that $V^{K_{0, \mathrm{fin}}}(\rho_{\infty})$ is isomorphic to a finite product of spaces of automorphic forms on $\mathrm{GL}(2, \mathbb{R})^{+}$, each of which is finite dimensional by Theorem \ref{autofin}. By strong approximation theorem, $\mathrm{GL}(2, \mathbb{A}) =\mathrm{GL}(2,\mathbb{Q})\mathrm{GL}(2, \mathbb{R})^{+}K_\mathrm{fin}$ and since $[K_{\mathrm{fin}}:K_{0, \mathrm{fin}}] < \infty$, the set of double cosets $$\mathrm{GL}(2, \mathbb{Q}) \mathrm{GL}(2, \mathbb{R})^{+} \backslash\mathrm{GL}(2, \mathbb{A})/K_{0, \mathrm{fin}}$$ is finite. Let $g_1, \dots, g_n$ be a set of representatives - we may assume $g_{i, \infty} = 1$ for all $i$. For $\phi\in V^{K_{0, \mathrm{fin}}}$, we can associate $h$ functions $\Phi_i$ on $\mathrm{GL}(2, \mathbb{R})^{+}$ defined by $\Phi_{i}(g_\infty) = \phi(g_{\infty}g_{i})$. Any $g\in \mathrm{GL}(2, \mathbb{A})$ can be written as $g = \gamma g_{\infty}g_{i}k_{0}$ for $\gamma\in \mathrm{GL}(2, \mathbb{Q}), g_{\infty} \in \mathrm{GL}(2, \mathbb{R})^{+}$ and $k_{0} \in K_{0, \mathrm{fin}}$, so $\phi(g) = \Phi_{i}(g_{\infty})$. This means that $\phi$ is uniquely determined by $\Phi_i$’s, so it is sufficient to show that each of these lies in a finite dimensional vector spaces.

Let $\Gamma$ be the projection onto $\mathrm{GL}(2, \mathbb{R})^{+}$ of $\mathrm{GL}(2, \mathbb{Q}) \cap (\mathrm{GL}(2, \mathbb{R})^{+}K_{0, \mathrm{fin}})$, which became a finite index subgroup of $\mathrm{SL}(2, \mathbb{Z})$ (in fact, this is a congruence subgroup). Then it is easy to check that $\Phi_i \in \mathcal{A}(\Gamma \backslash\mathrm{GL}(2, \mathbb{R})^{+}, 1, \omega_{\infty})$ where $\omega = \prod_{v} \omega_{v}$. (Moderate growth of $\Phi_i$ follows from that of $\phi$.) Since $\rho_{\infty}$-isotypic subspace of this space is finite dimensional by Theorem \ref{autofin}, so is $V^{K_{0, \mathrm{fin}}}(\rho_\infty)$. ◻

Proof. By no small subgroup argument, $\ker \rho$ contains an open subgroup of $K_{\mathrm{fin}}$ and so there exists a finite set of places $S$ containing $S_{\infty}$ such that $\rho(K_{v}) = 1$ for all $v\not\in S$. Then $\rho$ factors through the projection $$K \twoheadrightarrow K / (\prod_{v\not\in S} K_{v}) \simeq \prod_{v\in S} K_{v}$$ where the last group is a finite direct product of compact groups. Any irreducible representation of such group has a form $\otimes_{v\in S} \rho_{v}$ where $\rho_{v}$ is an irreducible representation of $K_{v}$. Then $\rho$ is isomorphic to the restricted tensor product with $(\rho_{v}, V_{v})$ trivial (one-dimensional) for $v\not\in S$. ◻

Proof. Since $\Lambda$ is nonzero, there exists a nonzero pure tensor $\xi^{\circ} = \otimes_{v} \xi_{v}^{\circ}$ such that $\Lambda(\xi^{\circ})= 1$. (Note that changing finite number of $\xi_{v}^{\circ}$ does not change the restricted tensor product.) For each place $w$ of $F$, let $i_{w} : V_{w} \to V, \xi_{w} \mapsto \xi_{w} \otimes (\bigotimes_{v\neq w} \xi_{v}^{\circ})$ and $\Lambda_{w} = \Lambda \circ i_{w}$. Then $\Lambda_{w}(\xi_{w}^{\circ}) =1$ and $\Lambda_{w}$ is a Whittaker functional on $V_{w}$. To prove the equation, we can use induction on the cardinality of the finite set $S$ such that $\xi_{v} =\xi_{v}^{\circ}$ for all $v\not\in S$. Base case is trivial because both sides are 1, and to add a single place $w$ to $S$, let’s assume that $\xi_{v} = \xi_{v}^{\circ}$ for all $v\not\in S\cup \{w\}$. Then $$x_{w} \mapsto \Lambda \left( x_{w} \otimes \left( \bigotimes_{v\neq w} \xi_{v}\right)\right)$$ is a Whittaker functional on $V_{w}$, so the uniqueness implies that there exists $c\in \mathbb{C}$ such that $$\Lambda \left( x_{w} \otimes \left( \bigotimes_{v\neq w} \xi_{v}\right)\right) = c\Lambda_{w} (x_{w}).$$ Now evaluate at $x_{w} = \xi_{w}^{\circ}$ and it gives a result for $S \cup \{w\}$. Uniqueness follows from the uniqueness of local Whittaker functionals. ◻

Proof. Let $F$ be a non-archimedean and $0\neq W_{\xi}\in \mathcal{W}$, so $W_{\xi}(g_0)\neq 0$ for some $g_0\in G$. For given $\mathcal{H}_G$-action, one can prove that there exists a representation $\pi$ of $\mathrm{GL}(2, F)$ such that the corresponding $\mathcal{H}_G$-action coincides with the previous one. If we denote the right translation action by $\rho$, then $W_{\pi(g)\xi} = \rho(g)W_{\xi}$ for $g\in \mathrm{GL}(2, F)$, so that $W_{\pi(g_{0})\xi} (1) = W_{\xi}(g_0)\neq 0$. If $\pi$ is spherical and the conductor of $\psi_v$ is $\mathcal{O}_v$, then by Theorem \ref{sphps}, $\pi$ is a spherical principal series representation. Then the nonvanishing of $W_{0}(1)$ follows from the explicit formula Theorem \ref{explicitsph}.

If $F$ is archimedean, we have to be more careful. Since $V$ is an admissible $\mathcal{H}_G$-module, it is also a $(\mathfrak{g}, K)$-module, and we can make use of the action $\pi : K\to \mathrm{End}(V)$. Since $K$ intersects every connected component of $G$, by applying $\pi(k)$ for suitable $k$ we may assume $W_{\xi}$ does not vanish identically on the connected component of the identity of $G$. Since $W_{\xi}$ is analytic (this is a solution of certain 2nd order differential equation), so $DW_{\xi}(1)\neq 0$ for some $D\in U\mathfrak{g}$. This equals $W_{D\xi}(1)$, so we may take $W = W_{D\xi}$. ◻

Proof. First, assume that each $\pi_v$ has a Whittaker model $\mathcal{W}_v$. By the previous proposition, if $\pi_v$ is a spherical representation and the conductor of $\psi_v$ is $\mathcal{O}_v$, there exists $W_v^{\circ} \in \mathcal{W}_v$ such that $W_{v}^{\circ}(k_v) = 1$ for all $k_v\in \mathrm{GL}(2, \mathcal{O}_v)$, which is a spherical element of $\mathcal{W}_v$. Then we can define a global Whittaker model $\mathcal{W}$ as the space of all finite linear combinations of functions of the form $W_{\xi}$ where $$W_{\xi}(g) = \prod_{v} W_{v, \xi_v} (g_v), \qquad g = (g_v) \in \mathrm{GL}(2, \mathbb{A})$$ where $\xi = \otimes_v \xi_v$ is a pure tensor in $V = \otimes_v V_v$ (so that $\xi_v = \xi_v^{\circ}$ for almost all $v$) and $W_{v, \xi_v^{\circ}} = W_{v}^{\circ}$ for almost all $v$. Then the product is well-defined because $W_{v, \xi_v}(g_v) =1$ for almost all $v$ and $\mathcal{W}$ affords an irreducible admissible representation of $\mathrm{GL}(2, \mathbb{A})$. Rapid decay of $W_{\xi}$ follows from the local results - see Theorem \ref{archwit} and \ref{pskr}.

Uniqueness proof is similar to the proof of Theorem \ref{ffwituniq}. We will show that if $\pi$ has a Whittaker model $\mathcal{W}$, then it is same as the one just described. Let $\xi\mapsto W_{\xi}$ be an isomorphism $V\simeq \mathcal{W}$.

First, one can show that there exists $\xi\in V$ such that $W_{\xi}(1) \neq 0$. The argument is almost same as the proof of the previous proposition. So we may assume $\xi$ is a pure tensor and $W_{\xi}(1) = 1$. Let $\xi = \otimes_v \xi_v^{\circ}$ where $\xi_v^{\circ}$ is $\mathrm{GL}(2, \mathcal{O}_v)$-fixed for almost all $v$.

Now for each $v$, if $\xi_v\in V_v$ and $g_v\in \mathrm{GL}(2, F_v)$ we define $$W_{v, \xi_v}(g_v) = W_{i_v(\xi_v)}(g_v)$$ where $i_v:V_v \to V$ is the map in the proof of Theorem \ref{ffwituniq}. Then the space of functions $W_{v, \xi_v}$ form a Whittaker model $\mathcal{W}_v$ for $\pi_v$ and $W_{v, \xi_{v}^{\circ}}(1) = 1$. Let $S$ be a finite set of places and let $\mathbb{A}_{S} = \prod_{v\in S} F_{v} \subset \mathbb{A}$. By induction on the size of $S$, we can prove that $$W_{\xi}(g) = \prod_{v\in S} W_{v, \xi_v}(g_v)$$ for all $g = (g_v) \in \mathrm{GL}(2, \mathbb{A}_S)$. Now for an arbitrary $g = (g_v) \in \mathrm{GL}(2, \mathbb{A})$ and an arbitrary pure tensor $\xi = \otimes_v \xi_v \in V$, there exists a finite set $S$ of places such that if $v\not\in S$, then $v$ is non-archimedean, $\xi_v$ is $K_v$-fixed, and $g_v\in K_v$. Let $h \in \mathrm{GL}(2, \mathbb{A}_\mathrm{fin})$ be the adele such that $h_v = g_v$ for $v\in S$ and $h_v = 1$ for $v\not\in S$. Then $$W_{\xi}(g) = W_{\xi}(h) = \prod_{v\in S} W_{v, \xi_v}(h_v) = \prod_{v} W_{v, \xi_v}(g_v),$$ so we get the exactly same equation as before. This proves the global uniqueness. ◻

Proof. It is not hard to see that $W_{\phi}$ satisfy the transformation law and of moderate growth. Also, $(\mathfrak{g}_\infty, K_\infty)$ and $\mathrm{GL}(2, \mathbb{A}_{\mathrm{fin}})$ act on $\mathcal{W}$ by right translation, and the action is compatible with the action on $V$. So we only need to show that the map $\phi\mapsto W_{\phi}$ is injective, and this will follow from the Fourier expansion.

To prove the Fourier expansion, let $f:\mathbb{A}\to \mathbb{C}$ be a function $$f(x) = \phi\left( \begin{pmatrix} 1 & x \\ & 1 \end{pmatrix} g\right).$$ Then $f$ is continuous (since $\phi$ is) and $f(x+a) = f(x)$ for all $a\in F$. So it can be regarded as a function on the compact group $\mathbb{A}/F$, and therefore it has a Fourier expansion as $$\phi\left( \begin{pmatrix} 1 & x \\ & 1 \end{pmatrix} g\right) = \sum_{\alpha\in F} C(\alpha) \psi(\alpha x)$$ with $$C(\alpha) = \int\displaylimits_{\mathbb{A}/F} \phi\left( \begin{pmatrix} 1 & x \\ & 1 \end{pmatrix} g\right) \psi(-\alpha x)dx.$$ We have $C(0) = 0$ since $\phi$ is cuspidal, and for $\alpha \neq 0$, because $\phi$ is automorphic \begin{align*}C(\alpha) &= \int\displaylimits_{\mathbb{A}/F} \phi\left( \begin{pmatrix} \alpha & \\ & 1 \end{pmatrix} \begin{pmatrix} 1 & x \\ & 1 \end{pmatrix} g\right) \psi(-\alpha x) dx \\ &= \int\displaylimits_{\mathbb{A}/F} \phi\left( \begin{pmatrix} 1 & \alpha x \\ & 1 \end{pmatrix} \begin{pmatrix} \alpha & \\ & 1 \end{pmatrix} g\right) \psi(-\alpha x) dx\end{align*} and the change of variables $x\mapsto \alpha^{-1} x$ gives $$C(\alpha) = W_{\phi}\left(\begin{pmatrix} \alpha & \\ & 1 \end{pmatrix} g\right).$$ Now put $x = 0$ and we obtain the equation. ◻

proof of Theorem \ref{multone} when $n=2$. First, we prove weaker form of the result (weak multiplicity one theorem), which is that if two automorphic representation $(\pi, V)$ and $(\pi, V')$ satisfies $\pi_{v}\simeq \pi'_{v}$ for all $v$, then $V = V'$. By the existence theorem, if $\mathcal{W}$ is the Whittaker model of $\pi$ then $V$ consists of the space of all functions $\phi$ of the form $$\phi(g) = \sum_{\alpha\in F^{\times}} W\left( \begin{pmatrix} \alpha & \\ & 1 \end{pmatrix}g \right), \quad W\in \cal W$$ and by the same reasoning, $V'$ consists of the same space. So we get $V = V'$.

For the original statement, let $(\pi, V)$ and $(\pi', V')$ be cuspidal automorphic representations such that such that $\pi \simeq \otimes_{v}\pi_{v}, \pi'\simeq \otimes_{v}\pi_{v}'$ and $\pi_{v}\simeq \pi_{v}'$ for all $v\not\in S$, where $S$ is a finite set of non-archimedean places. Let $\mathcal{W}_{v}$ and $\mathcal{W}_{v}'$ be the Whittaker models of $\pi_{v}$ and $\pi_{v}'$. For each $v$, we choose a nonzero $W_{v}\in \mathcal{W}_{v}$ so that

• $W_{v}(k_{v}) = 1$ for all $k_{v}\in K_{v}$, for all but finitely many $v$,

• $W_{v}\left(\begin{smallmatrix} y & \\ & 1 \end{smallmatrix}\right)$ on $F_{v}^{\times}$ is compactly supported for $v\in S$.

(First choice is possible because $\pi_v$ is spherical all but finitely many $v$. The second assertion follows from Theorem \ref{kicpt} - for any $\sigma \in C_{c}^{\infty}(F_{v}^{\times})$, there exists $W_v \in \mathcal{W}_v$ such that $\sigma(y) = W_{v} \left(\begin{smallmatrix} y & \\ & 1 \end{smallmatrix}\right)$.) Then chose $W_{v}'\in \mathcal{W}_{v}'$ as follows. For $v\not\in S$, $\mathcal{W}_{v}' = \mathcal{W}_{v}$ so choose $W_{v}' = W_{v}$. If $v\in S$, then by the Theorem \ref{kicpt} we may arrange that $$W_{v}\begin{pmatrix} y & \\ & 1 \end{pmatrix} = W_{v}'\begin{pmatrix} y & \\ & 1 \end{pmatrix}$$ for all $y\in F_{v}^{\times}$. With this choices, define $\phi\in V$ by $$\phi(g) = \sum_{\alpha\in F^{\times}} W\left(\begin{pmatrix} \alpha & \\ & 1 \end{pmatrix}g \right),$$ where $g = (g_{v})\in \mathrm{GL}(2, \mathbb{A}), W(g) = \prod_{v} W_{v}(g_{v})$, and similarly $\phi'\in V'$ from $W_{v}'$. Then it is enough to show that $\phi = \phi'$. By definition, $\phi(g) = \phi'(g)$ for all $g = \left(\begin{smallmatrix} y & \\ & 1 \end{smallmatrix}\right)$ where $y\in \mathbb{A}^{\times}$. Also, $W_{v}= W_{v}'$ if $v$ is archimedean and $W, W'$ are right invariant uner some open subgroup $K_{0}$ of $\mathrm{GL}(2, \mathbb{A}_{\mathrm{fin}})$, and they are automorphic. So $\phi(g) = \phi'(g)$ for $g = \gamma\left(\begin{smallmatrix} y & \\ & 1 \end{smallmatrix}\right) g_{\infty}k_{0}$, where $\gamma\in \mathrm{GL}(2, F), y\in \mathbb{A}^{\times}, g_{\infty}\in \mathrm{GL}(2, F_{\infty}), k_{0}\in K_{0}$, and the strong approximation concludes that $\phi = \phi'$. Since $W$ is nonzero, $\phi$ is nonzero and we can express $W$ in terms of $\phi$. So $V\cap V'\neq\emptyset$ and this proves $V = V'$. ◻

Proof. Since $\phi$ is automorphic, we have \begin{align*}Z(s, \phi, \xi) &= \int\displaylimits_{\mathbb{A}^{\times}/F^{\times}} \left( w_1 \begin{pmatrix} y & \\ & 1 \end{pmatrix} \right) |y|^{s-1/2} \xi(y)d^{\times} y \\ &= \int\displaylimits_{\mathbb{A}^{\times}/F^{\times}} \phi\left( \begin{pmatrix} 1 & \\ & y \end{pmatrix} w_{1}\right) |y|^{s-1/2} \xi(y) d^{\times} y.\end{align*} If we substitute $y^{-1}$ for $y$, we obtain $$\int\displaylimits_{\mathbb{A}^{\times}/F^{\times}} (\pi(w_1)\phi)\begin{pmatrix} y & \\ & 1 \end{pmatrix} |y|^{-s+1/2} (\xi \omega)^{-1}(y)d^{\times}y = Z(1-s, \pi(w_1)\phi, \xi^{-1}\omega^{-1}).$$ ◻

Proof. The proof uses explicit formula of $W_{v}$ in Theorem \ref{explicitsph} in terms of the Satake parameters $\alpha_{1}, \alpha_{2}$. Recall that $$W_{v}\begin{pmatrix} y & \\ & 1 \end{pmatrix} = \begin{cases} q^{-m/2} \frac{\alpha_{1}^{m+1} - \alpha_{2}^{m+1}}{\alpha_{1} - \alpha_{2}} & m\geq 0 \\ 0 & \text{otherwise.} \end{cases}$$ where $m = \mathrm{ord}_{v}(y)$ and $q = q_{v} = |\mathcal{O}_{v}/(\varpi_{v})|$. (Here the formula is slightly different because we use the different normalization $W_{v}(1) = 1$.) We can break the integral up into a sum over $m=0$ to $\infty$ to obtain \begin{align*}&\sum_{m=0}^{\infty} q^{-m/2} \frac{\alpha_{1}^{m+1} - \alpha_{2}^{m+1}}{\alpha_{1} - \alpha_{2}} q^{m/2 - ms}\xi_{v}(\varpi)^{m} \\ &= \frac{1}{\alpha_{1} - \alpha_{2}} \left( \frac{\alpha_{1}}{1-\alpha_{1}\xi_{v}(\varpi_{v})q^{-s}} - \frac{\alpha_{2}}{1-\alpha_{2}\xi_{v}(\varpi_{v})q^{-s}}\right)\\ &= L_{v}(s, \pi_{v}, \xi_{v}).\end{align*} ◻

Proof. Choose a pure tensor $\phi = \otimes_{v}\phi_{v}\in V$ such that $\phi_{v}$ is spherical for $v\not\in S$ and normalized so that if $W_{v}$ is a local Whittaker function corresponding to $\phi_{v}$, then $W_{v}(1) =1$ for $v\not\in S$. We will evaluate $$\left( \prod_{v\in S} Z_{v}(s, W_{v}, \xi_{v})^{-1} \right) Z(s, \phi, \xi)$$ in two different ways. First, it is easy to check that this equals the LHS of the above functional equation for large $\Re s$. Now, take $-\Re s$ to be large and positive. Local functional equation allow us to write above equation as \begin{align*}&\left(\prod_{v\in S} Z_{v}(s, W_{v}, \xi_{v})^{-1}Z_{v}(1-s, \pi(w_{1})W_{v}, \xi_{v}^{-1}\omega_{v}^{-1})\right) \\ &\times \prod_{v\not\in S} Z_{v}(1-s, \pi(w_{1})W_{v}, \xi_{v}^{-1}\omega_{v}^{-1}). \end{align*} Thus we will obtain the RHS if we show $$Z_{v}(s, \pi(w_{1})W_{v}, \xi_{v}^{-1}\omega_{v}^{-1}) = L_{v}(s, \widehat{\pi}_{v}, \xi_{v}^{-1})$$ for $v\not\in S$. Since $v$ is unramified, $W_{v}$ is the spherical vector and \begin{align*}Z_{v}(s, \pi(w_{1})W_{v}, \xi_{v}^{-1}\omega_{v}^{-1}) &= Z_{v}(s, W_{v}, \xi_{v}^{-1}\omega_{v}^{-1})\\ &= L_{v}(s, \pi_{v}, \omega_{v}^{-1}\xi_{v}^{-1}) \\ &= L_{v}(s, \widehat{\pi}_{v}, \xi_{v}^{-1}).\end{align*} The last equality follows from $\widehat{\pi}_{v} \simeq \omega_{v}^{-1} \otimes \pi_{v}$, or by direct calculation (if $\alpha_1, \alpha_2$ are Satake parameters of $\pi_v$, then $\alpha_1^{-1}, \alpha_{2}^{-1}$ are Satake parameters of $\widehat{\pi}_v$). ◻

Proof. By Theorem \ref{cuspdecomad}, $L^{2}_{0}$ decomposes as Hilbert space direct sum of irreducible invariant subspaces. Now choose $(\pi, V) \subseteq L_{0}^{2}$ such that the projection of $\phi$ to $V$ is nonzero. We will show that $\pi$ is uniquely determined by eigenvalues of $T_{p}$ with $p\nmid N$ and $\omega$. This will show that $\phi \in \pi$. Note that $\phi$ is $K_{v} = \mathrm{GL}(2, \mathbb{Z}_{p})$-fixed for $p\nmid N$ since $\lambda$ is trivial on that group.

For $p\nmid N$, let $G_{p} = \mathrm{GL}(2, \mathbb{Q}_{p})$, $\mathcal{H}_{p} =C_{c}^{\infty}(G_p)$ be the Hecke algebra and $\mathcal{H}_{p}^{\circ} = C_{c}^{\infty}(K_{p} \backslash G_{p} / K_p)$ be the spherical Hecke algebra. We studied the structure of spherical Hecke algebra in Section 3.7 - $\mathcal{H}_p^{\circ}$ is commutative and generated by three elements $T(\mathfrak{p}), R(\mathfrak{p}),$ and $R(\mathfrak{p})^{-1}$. (See Theorem \ref{nonarchsphcom}, Proposition \ref{heckerel} and Proposition \ref{heckegen}.) We will use new notations $\mathbb{T}_{p} := T(\mathfrak{p})$ and $\mathbb{R}_{p} := R(\mathfrak{p})$ in here to avoid confusion between classical and adélized Hecke operators. So we have $$\mathbb{T}_{p} := \mathds{1}_{K_{p} \left(\begin{smallmatrix} \varpi_p & \\ & 1 \end{smallmatrix}\right) K_{p}}, \quad \mathbb{R}_{p} := \mathds{1}_{K_{p} \left(\begin{smallmatrix} \varpi_{p} & \\ & \varpi_{p} \end{smallmatrix}\right)}, \quad \mathbb{R}_{p} := \mathds{1}_{K_{p} \left(\begin{smallmatrix} \varpi_{p} & \\ & \varpi_p \end{smallmatrix}\right)^{-1}}$$ where $\varpi_{p}$ is the idele whose $p$th component is $p$ and all of whose other components are 1. As before, we can decompose the double coset as $$K_{p} \begin{pmatrix} \varpi_p & \\ & 1 \end{pmatrix} K_{p} = \bigcup_{i=1}^{p+1} i_{p}(\xi_{i})K_p$$ where $i_{p}: \mathrm{GL}(2, \mathbb{Q}) \to \mathrm{GL}(2, \mathbb{A})$ is the map induced by the composition $\mathbb{Q}\hookrightarrow \mathbb{Q}_p \hookrightarrow \mathbb{A}$ and $$\xi_{i} = \begin{pmatrix} p & i \\ & 1 \end{pmatrix} \quad (1\leq i\leq p), \quad \xi_{p+1} = \begin{pmatrix} 1 & \\ & p \end{pmatrix}.$$ Also, we have an action $\rho$ of $\mathcal{H}_{p}$ on automorphic forms given by $$(\rho(\sigma)\phi)(g) = \int\displaylimits_{\mathrm{GL}(2, \mathbb{A})} \sigma(h)\phi(gh)dh, \quad \sigma\in \mathcal{H}_p$$ and we will denote $\rho(\mathbb{T}_{p})\phi$ as $\mathbb{T}_{p}(\phi)$. We will evaluate $(\mathbb{T}_{p}\phi)(g)$ when $g = \gamma g_{\infty} k_{0}$. Since $\phi$ is right $K_{p}$-invariant, we have $$(\mathbb{T}_{p}\phi)(g) = \sum_{i=1}^{p+1} \phi(gi_{p}(\xi_{i})).$$ For each $1\leq i\leq p+1$ and $k_{0} \in K_{0}(N)$, there exists $1\leq j\leq p+1$ and $k_{0}'\in K_{0}(N)$ such that $k_{0}i_{p}(\xi_{i}) = i_{p}(\xi_{j})k_{0}'$. If we write $\xi_{j} = \xi_{j, \infty}\xi_{j, \mathrm{fin}}$, then $$gi_{p}(\xi_{i}) = (\gamma\xi_{j}) (\xi_{j, \infty}^{-1}g_{\infty}) (\xi_{j, \mathrm{fin}}^{-1}i_{p}(\xi_{j})k_{0}')$$ where each component lies in $\mathrm{GL}(2, \mathbb{Q}), \mathrm{GL}(2, \mathbb{R})^{+}$ and $K_{0}(N)$. (Note that $\xi_{j}$ is considered as an element of $\mathrm{GL}(2, \mathbb{A})$ via diagonal map $\mathrm{GL}(2, \mathbb{Q})\hookrightarrow \mathrm{GL}(2, \mathbb{A})$, and the $p$th component of $\xi_{j, \mathrm{fin}}^{-1}i_{p}(\xi_j)$ is 1.) Then $$\phi(g i_{p}(\xi_{i})) = F(\xi_{j, \infty}^{-1}g_{\infty}) \lambda(\xi_{j, \mathrm{fin}}^{-1} i_{p}(\xi_j) k_{0}').$$ We know that $\lambda$ is determined by places $v|N$ and for each $v$, the $v$th component of $i_{p}(\xi_{j})k_{0}'$ is the same as the $v$th component of $k_{0}'$ or $k_{0}$. Thus we get $$\phi(gi_{p}(\xi_{i})) = (F|_{\chi}\xi_{j, \infty}^{-1})(g_{\infty})\lambda(k_{0})$$ and so if $F$ is an eigenfunction of the classical Hecke operator $T_{p} = T_{\left(\begin{smallmatrix} p & \\ & 1 \end{smallmatrix}\right)}$, then $\phi$ is an eigenfunction of $\mathbb{T}_p$ with the same eigenvalue. Similarly, we have $$(\mathbb{R}_{p}\phi)(g) = \int\displaylimits_{K_{p}\left(\begin{smallmatrix} \varpi_{p} & \\ & \varpi_{p} \end{smallmatrix}\right)} \phi(gh)dh = \phi(g) \omega(\varpi_{p}) = \chi(p)\phi(g)$$ so $\phi$ is an eigenfunction of $\mathbb{R}_{p}$ with eigenvalue $\chi(p)$. To summarize, $\phi$ is an eigenvector of $\mathcal{H}_{p}$, and the eigenvalues are determined by $\chi$ and the eigenvalues of the classical Hecke operators on $F$.

The projection of $L_{0}^{2}$ onto the invariant subspace $V$ is $\mathrm{GL}(2, \mathbb{A})$-equivariant, so the image of $\phi$ in $V$ is an eigenvector of $\mathcal{H}_p$ with the same eigenvalues. Now Theorem \ref{sphchar} tells us that this determines the irreducible constituent $\pi_{p}$ of $\pi$, and so $\pi$ is itself determined by the global multiplicity one theorem. ◻

Proof. By adélizing it, we obtain two automorphic representations $\pi_{f}, \pi_{g}$ of $\mathrm{GL}(2, \mathbb{A})$. Since these are Hecke eigenforms, we have $T_{p}f = a_{p}f$ and $T_{p}g = b_{p}f$ for all prime $p$, and $a_{p}$, $b_{p}$ are also eigenvalues of $\mathbb{T}_{p} \in \mathcal{H}_{p}$ for each components of representations $\pi_{f, p}$, $\pi_{g, p}$. Since $\mathbb{R}_{p}$ acts trivially, Theorem \ref{sphchar} implies $\pi_{f, p} \simeq \pi_{g, p}$ for any $p$ with $a_{p}= b_{p}$. Now the multiplicity one gives us $\pi_{f} = \pi_{g}$, which is $a_{p}= b_{p}$ for all $p$. So we get $f = g$. ◻